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

VersionsA only A + B A + B + Csame questions, different order

AnswersQuestions only With answer key Key only

New sheet

GCSE Higher sample Paper 1 (non-calculator)

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

MathsUKwww.geekhero.co.uk

1.Work out 4368 ÷ 12.
(a)346
(b)364
(c)304
(d)36.4
2.The first four terms of a sequence are 4, 9, 14, 19. Work out an expression, in terms of n, for the nth term.
(a)n + 5
(b)5n + 4
(c)5n
(d)5n − 1
3.Write the ratio 5 : 8 in the form 1 : n.
(a)1:0.6
(b)1:1.6
(c)1:1.5
(d)0.6:1
4.u is the column vector with top number 5 and bottom number 1. v is the column vector with top number 2 and bottom number 4. Work out u − v, giving your answer as a column vector in the form (top, bottom).
(a)(−3, 3)
(b)(3, −3)
(c)(3, 3)
(d)(7, 5)
5.On the probability scale from 0 (impossible) to 1 (certain), evens is exactly halfway. An event is 3/10 more likely than evens. Work out the probability of that event, giving your answer as a decimal.
(a)0.53
(b)1.30
(c)0.30
(d)0.80
6.A vet records the masses, m kg, of 30 dogs at a clinic in Preston: 0 < m ≤ 10 — 11 dogs, 10 < m ≤ 20 — 5 dogs, 20 < m ≤ 30 — 5 dogs, 30 < m ≤ 40 — 9 dogs. Work out an estimate for the mean mass, in kg, using the midpoint of each class interval.
(a)5 kg
(b)570 kg
(c)20 kg
(d)19 kg
7.A pizza is cut into 12 equal slices. Ben eats 5 slices and Mia eats 3 slices. What fraction of the pizza is left, giving your answer in its simplest form?
(a)1/3
(b)2/3
(c)1/2
(d)7/12
8.A rope of length L metres is cut into 6 equal pieces, and 4 metres is then removed from one piece. Write an expression, in metres, for the length of that piece after the cut.
(a)(L − 4)/6
(b)L/6 + 4
(c)4 − L/6
(d)L/6 − 4
9.Write £3.60 : £2.40 as a ratio in its simplest form.
(a)360 : 240
(b)36 : 24
(c)2 : 3
(d)3 : 2
10.In triangle ABC, AB = 6 cm, AC = 8 cm and angle BAC = 60°. Work out the exact area of triangle ABC, giving your answer in the form $a 3$ .
(a) $93$ cm²
(b) $243$ cm²
(c) $123$ cm²
(d) $163$ cm²
11.A fair six-sided dice is rolled 90 times. Work out how many more times you would expect it to land on a number less than 4 than on a 6.
(a)45
(b)36
(c)30
(d)15
12.Robert is converting 0.999... (with the 9s recurring forever) into a fraction. He lets x = 0.999... . Multiplying by 10 gives 10x = 9.999... . Subtracting x from 10x gives 9x = 9, so x = 1. Which statement correctly explains this result?
(a)Only approximately 1, since it never truly reaches 1.
(b)Only works because the recurring digit is 9.
(c)Exactly 1: the subtraction has no rounding at any step.
(d)Wrong: 10x minus x gives 9x = 8.999..., not 9.
13.A model rocket's height is given by h = −5t² + 20t, where h is in metres and t is in seconds. A student says the graph of h against t is n-shaped. Which statement gives the correct verdict on the SHAPE and the reason that settles it from the equation?
(a)Yes — the t² coefficient is negative, giving an n-shape.
(b)No — it is U-shaped, since height starts by increasing.
(c)No — the shape comes from the story, not the coefficient.
(d)Yes, but only because the rocket eventually lands.
14.A hiker's distance from the start of the trail, in kilometres, is plotted against time, in minutes, since she set off. At t = 45 minutes she stops to rest, so her distance from the start is neither increasing nor decreasing at that instant. Which statement about the tangent to the distance–time graph at t = 45 minutes is correct?
(a)The gradient equals 45, the time in minutes she stops.
(b)The tangent is vertical, so its gradient is undefined.
(c)The tangent is horizontal, so its gradient is 0.
(d)The gradient can't be found without her distance too.
15.A rectangle has vertices at (2, 1), (9, 1), (9, 5) and (2, 5). Work out the area of the rectangle.
(a)22
(b)28
(c)35
(d)63
16.Solve x² − x − 12 = 0.
(a)x = 4 or x = −3
(b)x = −4 or x = 3
(c)x = 12 or x = −1
(d)x = 4 or x = 3
17.The density of a liquid is 0.92 g/cm³. Work out the density of the liquid in kg/m³.
(a)9.2 kg/m³
(b)92 kg/m³
(c)920 kg/m³
(d)0.00092 kg/m³
18.Two straight paths cross at a junction in a park. One of the angles between them, measured with a protractor, is 76°. What is the size of the angle immediately next to it on a straight line?
(a)14
(b)104
(c)76
(d)90
19.Work out the value of (x − 4)/2 + 3 when x = 10
(a)0
(b)4.5
(c)6
(d)11
20.Which expression is equivalent to 0.5(4x + 6) − x?
(a)3x + 3
(b)2x + 3
(c)x + 3
(d)x + 6

Answer key

(b) 364 — Divide in stages using multiples of 12. 12 × 300 = 3600, leaving a remainder of 4368 − 3600 = 768. Then 12 × 64 = 768, so 4368 ÷ 12 = 300 + 64 = 364. Placing the decimal point as though dividing 436.8 by 12 gives 36.4. Transposing the last two digits of 364 gives 346. Working out 768 ÷ 12 as 4 instead of 64, losing the tens digit, and adding 300 + 4 gives 304. So 4368 ÷ 12 = 364.
(d) 5n − 1 — The common difference is 5 (9−4=5), so the expression starts 5n. To match the first term when n=1, 5×1+c=4, so c=−1: the nth term is 5n−1. A candidate who uses the first term itself as the constant, instead of first term minus the common difference, would write 5n+4 (giving 9, 14, 19, 24 — one term too high throughout). A candidate who omits the constant term altogether would write just 5n (giving 5, 10, 15, 20, not matching the sequence). A candidate who adds the common difference to n instead of multiplying would write n+5 (giving 6, 7, 8, 9, far too small).
(b) 1:1.6 — To write 5 : 8 in the form 1 : n, divide both parts by 5, the first number, so that it becomes 1: 5 ÷ 5 = 1 and 8 ÷ 5 = 1.6, giving 1 : 1.6. Dividing both parts by 8 instead gives 0.6 : 1 (5 ÷ 8 = 0.625, rounded to 0.6) — the first part is no longer 1, so this is not in the required form. Dividing 5 by 8 but writing the result after the 1 gives 1 : 0.6, which divides in the wrong direction: n must come from 8 ÷ 5, not 5 ÷ 8. A slip in the division 8 ÷ 5, rounding it to 1.5 instead of the correct 1.6, gives 1 : 1.5.
(b) (3, −3) — Subtracting column vectors means subtracting the top numbers and subtracting the bottom numbers, in the order given: top = 5 − 2 = 3, bottom = 1 − 4 = −3, giving (3, −3). A candidate who works out v − u instead of u − v, reversing the order, gets (−3, 3). A candidate who makes a sign error on the bottom number, treating 1 − 4 as 3 instead of −3, gets (3, 3). A candidate who adds the vectors instead of subtracting gets (7, 5). Because the question asks for u − v, not v − u, the correct answer is (3, −3).
(d) 0.80 — Evens is exactly halfway along the scale, at 0.5, and 3/10 written as a decimal is 0.3. A probability 3/10 higher than evens is 0.5 + 0.3 = 0.80. Giving 0.30 as the answer converts the increase but never adds it to the value of evens. Adding the increase to 1, the 'certain' end of the scale, instead of to evens gives 1 + 0.3 = 1.30, which cannot be a probability. Writing 3/10 as 0.03 instead of 0.3, a place-value slip, gives 0.5 + 0.03 = 0.53.
(d) 19 kg — Method: estimate the mean of grouped data by multiplying each class's midpoint by its frequency, adding the four totals, then dividing by the total frequency. Working: the midpoints are 5, 15, 25 and 35 kg. The weighted totals are 11 × 5 = 55, 5 × 15 = 75, 5 × 25 = 125 and 9 × 35 = 315, which add to 570. Dividing by the 30 dogs gives an estimate of 570 ÷ 30 = 19 kg. Giving 5 kg reads off the midpoint of the modal class, 0 < m ≤ 10, the class with the most dogs — but the class with the most dogs is not where the mean falls, and neither is a substitute for actually calculating it. Giving 20 kg averages the four midpoints, (5 + 15 + 25 + 35) ÷ 4, treating every class as equally likely and ignoring that far more dogs are in the lightest and heaviest classes than in the middle two. Giving 570 kg stops after finding the correct weighted total and forgets the final division by the 30 dogs. Always weight each midpoint by its own frequency, and always finish by dividing by the total frequency, not the number of classes.
(a) 1/3 — Method: find the total fraction eaten, then subtract it from the whole pizza. Working: together they eat 5/12 + 3/12 = 8/12, so the fraction left is 12/12 − 8/12 = 4/12 = 1/3. Answer: 1/3. 2/3 comes from giving the fraction eaten instead of the fraction left. 7/12 comes from only subtracting Ben's slices and forgetting Mia's. 1/2 comes from comparing the 4 slices left with the 8 slices eaten, 4/8, a part-to-part comparison instead of comparing with the whole pizza of 12 slices.
(d) L/6 − 4 — Method: find the length of one equal piece first (divide by 6), then apply the later change (subtract 4) to that piece. Working: one piece is L/6 metres; removing 4 metres from it gives L/6 − 4. Answer: L/6 − 4. L/6 + 4 comes from adding the 4 metres instead of removing it. (L − 4)/6 comes from removing the 4 metres from the whole rope before cutting it into pieces, the wrong order. 4 − L/6 comes from subtracting the piece length from 4 instead of the other way round.
(d) 3 : 2 — Convert both amounts to pence: £3.60 = 360p and £2.40 = 240p, giving the ratio 360 : 240. Divide both parts by their highest common factor, 120, to get 3 : 2. Giving 360 : 240 has not been simplified at all. Giving 2 : 3 swaps the order. Giving 36 : 24 has been divided by 10, which is a common factor but not the highest one, so it is not yet in simplest form.
(c) $12\sqrt{3}$ cm² — Use Area = 1/2 × AB × AC × sin(angle BAC) = 1/2 × 6 × 8 × sin 60°. Since sin 60° = √3⁄2, this is 1/2 × 6 × 8 × √3⁄2 = 24 × √3⁄2 = 12√3 cm². Using AB × AB instead of AB × AC (taking 6 × 6 rather than 6 × 8) gives 1/2 × 6 × 6 × sin 60° = 18 × √3⁄2 = 9√3 cm². Using AC × AC instead of AB × AC (taking 8 × 8 rather than 6 × 8) gives 1/2 × 8 × 8 × sin 60° = 32 × √3⁄2 = 16√3 cm². Leaving out the 1/2 from the area formula gives 6 × 8 × sin 60° = 48 × √3⁄2 = 24√3 cm².
(c) 30 — The numbers less than 4 are 1, 2 and 3, so the probability of that event is 3/6, and the expected count in 90 rolls is 90 × 3/6 = 45. The probability of rolling a 6 is 1/6, and the expected count is 90 × 1/6 = 15. The difference between the two expected counts is 45 − 15 = 30. A candidate who answers 45 has given the expected count for 'less than 4' only, forgetting to subtract the other expected count. A candidate who answers 15 has given the expected count for '6' only. A candidate who answers 36 has used a dice with 5 possible numbers instead of 6, giving 90 × 3/5 = 54 and 90 × 1/5 = 18, a difference of 36.
(c) Exactly 1: the subtraction has no rounding at any step. — 10x − x removes the recurring part completely, because the digits after the decimal point in 10x and in x are identical from the tenths place onward, so they cancel exactly: 9.999... − 0.999... = 9.000... = 9. Nothing was rounded to reach 9x = 9, so x = 1 is an exact equality, not an approximation, and the statement that the value is exactly 1, with no rounding at any step, is the correct one. Calling it only approximately 1, on the ground that a recurring decimal can never reach a whole number, misunderstands what the subtraction has just shown: the recurring tail cancels completely, leaving no gap to approximate away. Claiming the method only works because the recurring digit is 9 is also wrong — the same subtraction cancels the recurring part for any repeating digit, not just 9; it is the choice of multiplier (10, matching the one-digit repeat) that makes the cancellation exact, not the digit itself. Saying 10x minus x gives 8.999... rather than 9 misreads the subtraction: 9.999... − 0.999... has no digit to borrow from, since every decimal digit in the two numbers matches, so the result is exactly 9, not 8.999... .
(a) Yes — the t² coefficient is negative, giving an n-shape. — The coefficient of t² in h = −5t² + 20t is −5, which is negative, so the graph is n-shaped with a maximum point — this matches the physical story of the rocket rising then falling, but the shape itself is decided by the negative coefficient of t², not by the story alone. Saying the shape comes from the story rather than the coefficient gets the reasoning backwards — the algebra determines the shape, and the story happens to agree with it. Saying it is U-shaped because height starts by increasing confuses the early part of the curve with its overall shape; a U-shaped curve would mean the rocket's height eventually increases again forever, which does not happen here. Saying it is n-shaped only because the rocket lands treats a consequence of the shape as if it were the cause.
(c) The tangent is horizontal, so its gradient is 0. — Method: at any point where a distance–time graph is momentarily neither increasing nor decreasing, the tangent to the graph at that point is horizontal, and the gradient of a horizontal line is 0 — this is the instantaneous rate of change at that instant. Working: since the hiker's distance is neither increasing nor decreasing at t = 45 minutes, the tangent there is horizontal, so its gradient is 0. Claiming the tangent is vertical, with an undefined gradient, is the opposite of what the stem says: a vertical tangent would mean the distance was changing infinitely fast at that instant, not that it had stopped changing, and on a distance–time graph it cannot happen at all. Reading the gradient as 45, the time value given in the stem, mistakes a value used to LOCATE the point for the rate of change AT that point. Claiming the gradient cannot be found without also knowing the distance at t = 45 minutes overlooks that 'momentarily stationary' already tells you the rate of change directly, without needing to read any distance value at all. Whenever a stem tells you a quantity is momentarily not changing, that is telling you the instantaneous rate of change directly — it is 0, and no further data is needed to find it.
(b) 28 — The width of the rectangle is the difference in x-coordinates, 9 − 2 = 7, and the height is the difference in y-coordinates, 5 − 1 = 4. The area is width × height = 7 × 4 = 28. 22 comes from using the perimeter formula, 2 × (7 + 4), instead of the area formula. 35 comes from multiplying 7 by 5 instead of 4, misreading one of the y-coordinates. 63 comes from multiplying 9 by 7, using an x-coordinate instead of the height.
(a) x = 4 or x = −3 — Method: find two numbers that multiply to give −12 and add to give −1 — these are −4 and 3. So x² − x − 12 = (x − 4)(x + 3) = 0, giving x = 4 or x = −3. Distractor origins: x = −4 or x = 3 has the signs the wrong way round; x = 4 or x = 3 makes both roots positive, ignoring the sign of −12; x = 12 or x = −1 comes from reading off the coefficient and the constant directly instead of factorising.
(c) 920 kg/m³ — Convert each unit in turn. Mass: 1 g = 0.001 kg. Volume: 1 m³ = 100 × 100 × 100 = 1 000 000 cm³. So a density of 0.92 g per cm³ is 0.92 × 1 000 000 = 920 000 g in every cubic metre, and 920 000 g = 920 000 × 0.001 = 920 kg. The two conversions leave a single factor of 1 000 000 × 0.001 = 1000, so in one step multiply g/cm³ by 1000: 0.92 × 1000 = 920 kg/m³. Multiplying by 100 instead of 1000 gives 92 kg/m³, using the factor for 1 m² rather than 1 m³ of volume. Multiplying by 10 instead of 1000 gives 9.2 kg/m³, moving the decimal point one place for a conversion that moves it three. Dividing by 1000 instead of multiplying gives 0.00092 kg/m³, going the wrong way between the units — a kilogram is heavier than a gram, but a cubic metre is a million times bigger than a cubic centimetre, so the number must get larger, not smaller. The liquid's density is 920 kg/m³.
(b) 104 — Method: angles on a straight line add up to 180°. Working: 180° − 76° = 104°. A student who answers 76 has mistaken this for the vertically opposite angle, which is equal, rather than the adjacent angle on a straight line. A student who answers 90 has wrongly assumed the two paths must be perpendicular. A student who answers 14 has subtracted 76° from 90° instead of from 180°. Answer: 104°.
(c) 6 — Method: substitute the value, work out the top of the fraction first, then the division, and add the 3 last. Working: the top gives 10 − 4 = 6, dividing by 2 gives 6 ÷ 2 = 3, and adding 3 gives 3 + 3 = 6. Answer: 6. The distractors: 11 comes from dividing only the 4 by 2 instead of the whole of the top, giving 10 − 2 + 3; 4.5 comes from dividing the + 3 by 2 as well, giving (10 − 4 + 3) ÷ 2; 0 comes from subtracting the wrong way round on the top, giving (4 − 10) ÷ 2 = −3 and then −3 + 3.
(c) x + 3 — Expand the bracket: 0.5(4x + 6) = 2x + 3. Then subtract the x: 2x + 3 − x = x + 3. The option 2x + 3 comes from expanding the bracket correctly but then forgetting to subtract the x at all. The option x + 6 comes from forgetting to multiply the 6 inside the bracket by 0.5 (treating it as 2x + 6), then subtracting x. The option 3x + 3 comes from adding the x instead of subtracting it: 2x + 3 + x = 3x + 3.

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.