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
GCSE Higher sample Paper 1 (non-calculator)
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- 1.Simplify (5² × 2³)³ ÷ (5³ × 2²)²
- 2.The graph of y = x² − 5x + 6 crosses the x-axis at two points. By factorising, work out the x-coordinates of these two points.y = x² − 5x + 6
- 3.A train journey takes 2 hours 15 minutes. A bus journey to the same place takes 3 hours 45 minutes. Write the time for the train journey as a fraction of the time for the bus journey, giving your answer in its simplest form.
- 4.A student is proving that opposite angles of a cyclic quadrilateral ABCD sum to 180°, letting a = angle DAB and c = angle BCD, and using the fact that the angle at the centre is twice the angle at the circumference. She has already written: 'By the angle at the centre theorem, the reflex angle BOD equals 2a and the non-reflex angle BOD equals 2c.' Which statement must come immediately after this one in a correct proof?
- 5.A spinner can land on red, blue, green or yellow, and it cannot land on more than one colour. The probability that it lands on red is 0.05 and the probability that it lands on yellow is 0.35. The probability that it lands on blue is twice the probability that it lands on green. Work out the probability that it lands on green.
- 6.In a histogram of the lengths, x cm, of some rods, the bar for 10 ≤ x < 30 has a frequency density of 3 per cm. The bar for 30 ≤ x < 45 is twice as tall as the bar for 10 ≤ x < 30. Work out the number of rods with a length in the class 30 ≤ x < 45.
- 7.A café is planning its lunch menu. It will offer one soup from 3 choices and one sandwich from 4 choices, but the mushroom soup, one of the 3 soups, will only be served with the cheese sandwich, one of the 4 sandwiches, on a normal day. Work out how many different soup-and-sandwich combinations are available on a normal day.
- 8.Isla says that the lines y = 2x + 1 and y = −2x + 3 are parallel. Write down the statement that gives the correct answer for the correct reason.y = 2x + 1y = -2x + 3
- 9.Write £3.60 : £2.40 as a ratio in its simplest form.
- 10.A circle has centre O. Radii OA and OB are drawn, together with the arc AB, enclosing a sector. The straight chord AB is also drawn. Write down the name of the region enclosed between the chord AB and the arc AB.
- 11.A charity tombola has 30 tickets, 6 of which win a prize. Priya buys a ticket at random and does not return it. Her friend Tom then buys a second ticket at random from the remaining tickets. Work out the probability that both Priya and Tom win a prize.
- 12.Write these numbers in order, starting with the smallest: −1.4, 5/4, −6/5, 1.3, 0
- 13.Yusuf has £40 saved and adds £6 each week. His sister Aisha has £10 saved and adds £11 each week. After how many weeks will they have saved the same amount?
- 14.Grace's mean mark in her maths tests rises from 70 to 84. Work out the percentage increase in her mean mark.
- 15.Triangle A has a vertex at (1, 2). Triangle A is enlarged, centre (1, 1), to give triangle B, whose corresponding vertex is at (1, 5). Work out the scale factor of the enlargement.
- 16.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?
- 17.A map has a scale of 1:25000. A road on the map is 6 cm long. Work out the real-life length of the road, giving your answer in kilometres.
- 18.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 .
- 19.Solve the inequality 5 − x ≤ 2.
- 20.Expand and simplify 2(a + 6) − 2(a − 7)
Answer key
- (b) 2⁵ — Method: a power outside brackets multiplies the index of every factor inside them, and dividing powers of the same base subtracts their indices. Working: (5² × 2³)³ = 5⁶ × 2⁹ and (5³ × 2²)² = 5⁶ × 2⁴. Dividing gives 5⁶⁻⁶ × 2⁹⁻⁴, and since 5⁰ = 1 the whole expression reduces to 2⁵. Answer: 2⁵. The distractors: 2² comes from adding the outside index to each inside index instead of multiplying, which gives 5⁵ × 2⁶ over 5⁵ × 2⁴; 2¹³ comes from adding the indices of 2 when dividing, 9 + 4, instead of subtracting them; 2 comes from applying the outside power to the first factor inside each bracket only, leaving 5⁶ × 2³ over 5⁶ × 2².
- (c) x = 2 and x = 3 — x² − 5x + 6 factorises as (x − 2)(x − 3), since −2 and −3 multiply to give 6 and add to give −5. The graph crosses the x-axis where each bracket equals zero: x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. The option x = −2 and x = −3 comes from reading the signs inside the brackets directly instead of solving x − 2 = 0 and x − 3 = 0. The option x = 2 and x = −3 mixes up the sign of only one root. The option x = −1 and x = −6 comes from picking the wrong pair of factors of 6 (1 and 6 instead of 2 and 3) and then reading their signs directly from the brackets.
- (d) 3/5 — Convert both times to minutes: 2 hours 15 minutes = 135 minutes; 3 hours 45 minutes = 225 minutes. Put the train time over the bus time: 135/225. Divide both numbers by their highest common factor, 45: 135÷45 = 3, 225÷45 = 5, giving 3/5. (5/3 comes from writing the times the wrong way round. 2/5 comes from finding the difference, 225 − 135 = 90 minutes, and writing it as a fraction of the bus time, 90/225. 3/8 comes from comparing the train time to the total time for both journeys, 135/360.)
- (a) Reflex + non-reflex angle BOD = 360°, so 2a + 2c = 360°. — A correct proof must build in a strictly logical order: the reflex and non-reflex angles at O, which together make a complete turn about the centre, must be added to 360° before 2a and 2c can be combined and divided. The statement 'Reflex + non-reflex angle BOD = 360°, so 2a + 2c = 360°' is the only one that follows directly from having 2a and 2c already established, and it is exactly what is needed before the final division step. Jumping straight to 'Divide by 2 throughout to get a + c = 180°' skips the step that justifies why 2a + 2c equals 360° in the first place — there is nothing yet to divide. Repeating 'Angle BOD is 2a and 2c, by the centre theorem' does not move the proof forward at all, since that fact has already been established in the sentence given. Introducing 'OB = OD are radii, so triangle OBD is isosceles' brings in an unrelated triangle and an unrelated method that plays no part in this particular proof.
- (c) 0.2 — Let P(green) = x, so P(blue) = 2x. Red, blue, green and yellow are exhaustive: 0.05 + 0.35 + x + 2x = 1, so 0.4 + 3x = 1, giving 3x = 0.6 and x = 0.2. So P(green) = 0.2. Splitting the remaining 0.6 evenly between blue and green, ignoring the 2:1 ratio, gives 0.3. Working out x correctly but then reporting 2x, the probability of blue, gives 0.4. Stopping after finding that blue and green together account for 0.6, without dividing by the three equal shares of x, gives 0.6.
- (c) 90 — Method: the height of a bar is its frequency density, so twice as tall means twice the frequency density — not twice the frequency, because the two classes have different widths. Then frequency = frequency density × class width. Working: the first bar has frequency density 3 per cm, so the second has frequency density 2 × 3 = 6 per cm; the class 30 ≤ x < 45 is 45 − 30 = 15 cm wide, so its frequency is 6 × 15 = 90. Answer: 90 rods. The distractors: 120 comes from doubling the first bar's frequency instead of its height — the first class holds 3 × 20 = 60 rods, and doubling that ignores the fact that the second class is narrower; 45 comes from using the first bar's frequency density, 3, for the second bar, 3 × 15, and so never using the information that it is twice as tall; 6 comes from stopping at the frequency density of the taller bar and quoting a height as though it were a count.
- (d) 9 — Method: split into two cases — the soups with no restriction, and the mushroom soup on its own — then add the totals. Working: the 2 soups other than mushroom can be paired with any of the 4 sandwiches: 2 × 4 = 8. The mushroom soup can only be paired with the cheese sandwich: 1 combination. Total = 8 + 1 = 9. Answer: 9. 12 comes from working out 3 × 4 = 12 without applying the restriction at all. 8 comes from correctly finding the 2 unrestricted soups' 8 combinations, but forgetting to add back the 1 allowed mushroom-and-cheese combination. 11 comes from taking the unrestricted total of 12 and removing only 1 mushroom combination instead of all 3 disallowed ones.
- (b) No, because their gradients are 2 and −2 — Method: two lines are parallel exactly when their gradients are equal as signed numbers, so m is read from each equation written in the form y = mx + c and the two are compared. Working: y = 2x + 1 has gradient 2 and y = −2x + 3 has gradient −2; those are not equal, so the lines are not parallel, and indeed one slopes upwards while the other slopes downwards. Answer: No, because their gradients are 2 and −2. The distractors: saying yes because both gradients have size 2 comes from comparing the sizes of the gradients and ignoring their signs; saying yes because the gradients add to 0 comes from using a sum of zero as the test for parallel lines instead of equality of gradients; saying no because the y-intercepts are 1 and 3 reaches the right verdict by the wrong route, since the intercepts decide where the lines sit rather than whether they are parallel.
- (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.
- (b) segment — The region enclosed between a chord and the arc it cuts off is called a segment. A sector is the different region already named in the question, enclosed by the two radii and the arc, not by the chord. A tangent is a straight line that touches the circle at one point, not an enclosed region at all. The circumference is the total distance around the whole circle, a length, not a region.
- (a) 1/29 — The probability that Priya's ticket wins is 6/30. Since her ticket is not returned, there are now only 5 winning tickets left out of 29 tickets in total, so the probability that Tom's ticket also wins is 5/29. Multiplying these, 6/30 × 5/29 = 30/870 = 1/29. A candidate who answers 1/25 has treated Priya's ticket as returned, using 6/30 twice. A candidate who answers 1/30 has correctly reduced the winning tickets to 5 for Tom but forgotten to reduce the total number of tickets, using 5/30 instead of 5/29. A candidate who answers 11/59 has added the numerators and added the denominators, (6+5)/(30+29), instead of multiplying.
- (d) −1.4, −6/5, 0, 5/4, 1.3 — Method: convert the fractions 5/4 and −6/5 to decimals so every number is written the same way, then compare all five decimals. Working: 5/4 = 1.25 and −6/5 = −1.2. Comparing −1.4, −1.2, 0, 1.25 and 1.3 in size gives the order −1.4, −1.2, 0, 1.25, 1.3. Answer: −1.4, −6/5, 0, 5/4, 1.3. −6/5, −1.4, 0, 5/4, 1.3 swaps the two negative numbers, treating −6/5 as more negative than −1.4 even though −1.2 is closer to zero than −1.4. 1.3, 5/4, 0, −6/5, −1.4 lists the numbers from largest to smallest instead of smallest to largest. −1.4, −6/5, 0, 1.3, 5/4 swaps 5/4 and 1.3, comparing the numerator 5 directly with 1.3 instead of converting 5/4 to the decimal 1.25 first.
- (d) 6 — Method: write an expression for each person's savings after w weeks, and set them equal. Working: 40 + 6w = 10 + 11w. Subtract 6w from both sides: 40 = 10 + 5w. Subtract 10: 30 = 5w, so w = 6. Answer: 6 weeks. 0 comes from setting only the weekly amounts equal, 6w = 11w, and ignoring the different starting amounts entirely. 10 comes from adding the two starting amounts and dividing by the difference in weekly amounts, (40 + 10) ÷ (11 − 6), instead of forming and solving the correct equation. 1.76 comes from adding the two weekly amounts instead of subtracting them when rearranging, (40 − 10) ÷ (11 + 6).
- (a) 20% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is 84 − 70 = 14 marks, and 14 ÷ 70 = 0.2, so 0.2 × 100 = 20. Answer: an increase of 20%. The distractors: 14% comes from quoting the 14 mark increase as though marks and per cent were the same thing; 17% comes from dividing the 14 by the new mean 84 instead of by the original 70, which gives 17% to the nearest per cent; 120% is the new mean written as a percentage of the old one, which is the whole of the new mean rather than the increase.
- (b) 4 — The scale factor is the distance from the centre to the image, divided by the distance from the centre to the object: (5 − 1) ÷ (2 − 1) = 4 ÷ 1 = 4. (2.5 comes from dividing the raw y-coordinates, 5 ÷ 2, without first subtracting the centre's coordinate; 3 comes from subtracting the two distances instead of dividing them; 0.25 comes from dividing the distances the wrong way round.)
- (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) 1.5 km — Multiply the map length by the scale factor: 6 × 25000 = 150000 cm. Convert to kilometres, using 100 cm = 1 m and 1000 m = 1 km, so 100000 cm = 1 km: 150000 ÷ 100000 = 1.5 km. (1500 km comes from converting only as far as metres, 150000 ÷ 100 = 1500 m, and then writing kilometres on the end. 15 km comes from dividing by 10000 instead of 100000. 0.15 km comes from dividing by 1000000 instead of 100000.)
- (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) x ≥ 3 — Method: collect the number terms first; the x term is negative, so the final step multiplies both sides by −1, and that is the one step that turns the inequality sign round. Working: subtracting 5 from both sides of 5 − x ≤ 2 gives −x ≤ −3; multiplying both sides by −1 turns −x into x and −3 into 3, and because the multiplier is negative the ≤ becomes ≥, so x ≥ 3. Answer: x ≥ 3. The distractors: x ≤ 3 comes from multiplying by −1 without turning the sign round, the commonest slip on this type; x ≤ −3 comes from reading −x ≤ −3 as though the minus sign could simply be rubbed off the left-hand side; x ≥ −3 comes from turning the sign round correctly but leaving the right-hand side at −3 instead of multiplying it by −1 as well.
- (d) 26 — Method: expand both brackets, treating the second as multiplication by −2 so that both of its terms change sign, then collect like terms. Working: 2(a + 6) = 2a + 12 and −2(a − 7) = −2a + 14, so the expression becomes 2a + 12 − 2a + 14; the a terms give 2a − 2a = 0, so no term in a survives, and the numbers give 12 + 14 = 26. Answer: 26. The distractors: −2 comes from expanding the second bracket as −2a − 14, so that the numbers give 12 − 14; 4a − 2 comes from adding 2(a − 7) instead of subtracting it, giving 2a + 12 + 2a − 14; 13 comes from multiplying the 2 over only the first term of each bracket, giving 2a + 6 − 2a + 7.
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.