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.The decimal 0.272727... repeats the block 27 for ever. Write 0.27 recurring as a fraction in its simplest form.
- 2.The graph of y = sin x is transformed onto the graph of y = sin x + 1. Which statement correctly describes the transformation and the new range of the graph?y = sin(x)
- 3.A market stall sells apples at a fixed price per kilogram. Priya buys 3 kg for £5.40. Write the ratio of the mass in kilograms to the cost in pounds, in its simplest form.
- 4.Two traffic cones are geometrically similar. The smaller cone has height 6 m and the larger cone has height 9 m. Work out the ratio of the volume of the smaller cone to the volume of the larger cone, giving your answer in the form a : b in its simplest form.
- 5.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.
- 6.A scatter graph shows the midday temperature, x °C, and the number of ice creams sold at a seaside kiosk, y. The line of best fit is y = 3x − 20. Give a reason why the y-intercept of this line of best fit is not a sensible estimate of the number of ice creams sold.y = 3x − 20
- 7.Write 0.325 as a fraction in its simplest form.
- 8.Solve the simultaneous equations 5x − 2y = 16 and 3x + 2y = 16. Work out the value of x.
- 9.The ratio of cats to dogs in a shelter is 3:4. There are 18 cats. Work out the total number of animals in the shelter.
- 10.State whether the line segment joining A(−4, 3) and B(2, 6) is horizontal, vertical or neither, and give a reason for your answer.
- 11.A phone network sends automatic text alerts to customers. On average, 1,500 alerts are sent each day, and the probability that a customer replies 'STOP' to an alert is 0.18. Work out how many replies of 'STOP' the network should expect over a 30-day month.
- 12.Work out 1 1/2 ÷ 3/4 exactly, giving your answer in its simplest form.
- 13.Line P has equation y = −2x + 5 and line Q has equation 2y = x + 6. Are lines P and Q perpendicular to each other?y = -2x + 5y = x + 6
- 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.An isosceles triangle ABC has AB equal to AC. AM is a line from vertex A, perpendicular to BC, meeting BC at point M. Which condition proves that triangle ABM is congruent to triangle ACM?
- 16.A rectangular sheep pen has perimeter P, length l and width w, connected by the formula P = 2l + 2w. A farmer has 60 m of fencing for the perimeter and sets the width at 8 m. Work out the length of the pen.
- 17.Simplify the ratio 45 : 30 : 75 to its simplest form.
- 18.A conservatory has a roof panel shaped like a regular octagon. Work out the interior angle of the octagon, then use it to work out the size of the reflex angle at the same vertex, on the outside of the roof panel.
- 19.A coastguard radar at the origin covers a circular region modelled by x² + y² = 400, where each unit represents 1 kilometre. A boat travels along the straight line that touches the boundary of the region at the point (12, 16). Work out the equation of the line the boat travels along.
- 20.The diagram shows a distance–time graph for a cyclist travelling at a constant speed, for the first 6 minutes of a journey. Distance is in kilometres and time is in minutes. Work out how far the cyclist would travel in 20 minutes at the same speed.
Answer key
- (c) 3/11 — Method: let a letter stand for the recurring decimal, multiply by the power of ten that shifts exactly one repeating block past the point, subtract the original equation so that the recurring tail cancels, then solve and cancel. Working: let x = 0.272727...; the repeating block is two digits long, so multiply by 100 to give 100x = 27.272727...; subtracting gives 99x = 27, so x = 27/99; the highest common factor of 27 and 99 is 9, and 27 ÷ 9 = 3 with 99 ÷ 9 = 11. Answer: 3/11. The distractors: 27/100 comes from writing the repeating block over 100 instead of over 99, forgetting that subtracting x leaves 99x rather than 100x; 3/10 comes from rounding the decimal to one place and converting 0.3; 2/9 comes from treating only the 2 as recurring and converting 0.222... instead.
- (c) Translate +1 in y; new range 0 ≤ y ≤ 2 — Adding 1 to sin x translates the graph 1 unit in the positive y-direction, and every y-value on the range increases by 1: −1 + 1 = 0 and 1 + 1 = 2, giving a new range of 0 ≤ y ≤ 2. A translation in the x-direction would not add anything to the y-values, so 'Translate +1 in x; range unchanged' correctly leaves the range at −1 ≤ y ≤ 1 for that (wrong) transformation, but the transformation itself is not what y = sin x + 1 shows. 'Translate +1 in y; new range −2 ≤ y ≤ 0' correctly spots the translation but subtracts 1 from each bound instead of adding it. 'Reflect in the x-axis' mistakes the transformation for a reflection rather than a translation; reflecting sin x in the x-axis does leave the range at −1 ≤ y ≤ 1 unchanged in size, but that is not the transformation y = sin x + 1 actually applies.
- (c) 5 : 9 — Write the ratio mass : cost = 3 : 5.4. Multiply both parts by 10 to clear the decimal: 30 : 54. Both numbers share a factor of 6, so 30 ÷ 6 = 5 and 54 ÷ 6 = 9, giving 5 : 9. 3 : 5 comes from ignoring the decimal point and treating £5.40 as £5. 9 : 5 comes from writing the ratio the wrong way round, cost to mass instead of mass to cost. 1 : 18 comes from multiplying only the cost by 10 instead of both parts, giving 3 : 54, and then cancelling that correctly to 1 : 18 — the cancelling is fine, but the ratio being cancelled is not the right one.
- (b) 8 : 27 — The heights are in the ratio 6 : 9, which simplifies to 2 : 3. For similar solids, volume scales with the cube of the length ratio, so the volume ratio is 2³ : 3³ = 8 : 27. 2 : 3 is only the simplified length ratio, without cubing. 4 : 9 comes from squaring instead of cubing — that's the rule for areas, not volumes. 27 : 8 has the correct cubed values but in the wrong order, giving the larger cone's volume first instead of the smaller.
- (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.
- (a) x = 0 gives y = −20: a negative number sold — The y-intercept is the value the line predicts when x = 0: y = 3 × 0 − 20 = −20. A kiosk cannot sell a negative number of ice creams, so this is not a sensible estimate. The 3 in the equation is the gradient, not the intercept, so an option claiming x = 0 gives y = 3 has swapped the two numbers around — substituting x = 0 makes the 3x term equal 0, leaving −20, not 3. The danger of extrapolating to very high temperatures is a real issue with this line, but it is a different issue from the y-intercept, so it does not answer this question. And whether x = 0 could occur on a trading day is beside the point: the model still makes that prediction, and it is the prediction itself, −20, that is impossible.
- (a) 13/40 — Method: write the decimal over 1000 using its three decimal places, then simplify. Working: 0.325 = 325/1000 = 13/40 (dividing both numerator and denominator by 25). Answer: 13/40. 13/4 comes from writing the decimal over 100 instead of 1000, as if there were only two decimal places. 8/25 comes from rounding 0.325 down to 0.32 before converting. 3/8 comes from recalling the learned conversion 3/8 = 0.375 and matching it to 0.325 because both are three-place decimals beginning with 3, instead of converting the decimal given.
- (b) 4 — Adding the two equations: the y-terms, −2y and +2y, cancel, and the x-terms combine to 5x + 3x = 8x; the right-hand sides add to 16 + 16 = 32. This gives 8x = 32, so x = 4. A candidate who adds only one of the right-hand sides, instead of both, would get 8x = 16, so x = 2. A candidate who divides 32 by 4 instead of 8 would get x = 8. A candidate who subtracts the equations instead of adding them, getting 2x − 4y = 0, and then wrongly assumes y = 0, would get x = 0.
- (a) 42 — Find the value of one part: 18 ÷ 3 = 6. Find the number of dogs: 4 × 6 = 24. Add the cats and the dogs to find the total: 18 + 24 = 42. (24 is the number of dogs only, not the total number of animals. 126 comes from multiplying the number of cats by the total number of parts, 18 × 7, instead of finding one part first. 25 comes from adding the ratio numbers 3 and 4 directly to the number of cats.)
- (b) Neither, because both coordinates differ — A line segment is horizontal only when both points share the same y-coordinate, and vertical only when both points share the same x-coordinate. Here A has x-coordinate −4 and B has x-coordinate 2, which differ, and A has y-coordinate 3 and B has y-coordinate 6, which also differ, so the segment is neither horizontal nor vertical. Every point has a y-coordinate and an x-coordinate, so simply having one is not a reason for the line to be horizontal or vertical — both of those wrong reasons ignore that the coordinates must match, not just exist. The x-coordinates do increase from A to B, but an increasing x-coordinate on its own describes a slope, not a horizontal line.
- (b) 8,100 — First find the total number of alerts sent in the month: 1,500 × 30 = 45,000. Then apply the probability of a 'STOP' reply: 45,000 × 0.18 = 8,100. Stopping after finding only one day's expected replies, 1,500 × 0.18 = 270, forgets to scale up to the whole month. Multiplying the number of days by the probability instead of by the daily total of alerts gives 30 × 0.18 = 5.4, which rounds to 5. Shifting the decimal point in the probability, using 0.018 instead of 0.18, gives 45,000 × 0.018 = 810.
- (c) 2 — First write 1 1/2 as an improper fraction, 3/2. To divide by 3/4, multiply by its reciprocal, 4/3: 3/2 × 4/3 = 12/6 = 2. Dropping the whole number and dividing only the fractional part, 1/2 ÷ 3/4 = 1/2 × 4/3, gives 2/3. Multiplying by 3/4 directly instead of using its reciprocal, 3/2 × 3/4, gives 9/8. Using the reciprocal of the first fraction instead of the second, 2/3 × 3/4, gives 1/2.
- (a) Yes — the gradients multiply to −2 × 1/2 = −1. — Rearrange Q into the form y = mx + c: 2y = x + 6 gives y = (1/2)x + 3, so Q has gradient 1/2. P has gradient −2. Two lines are perpendicular exactly when the product of their gradients is −1: −2 × 1/2 = −1. Since this holds, P and Q are perpendicular. Distractor routes: "the product is −1, but perpendicular needs 1" works out the product correctly but misremembers the condition — the perpendicular test is a product of exactly −1, and parallel lines are spotted by their gradients being equal, not by a product of 1. "Q's gradient is 2, and −2 × 2 = −4" comes from reading the 2 in front of y in 2y = x + 6 as the gradient, instead of dividing the whole equation by 2 first to reach y = (1/2)x + 3, where the gradient is 1/2. "Both equations have a negative x-term" is not a valid test at all — P's equation does have a negative x-term, but Q's, once rearranged, does not, and matching signs say nothing about the actual gradients.
- (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.
- (d) RHS, using AM as common side — Triangle ABM and triangle ACM both have a right angle at M, since AM is perpendicular to BC. AB and AC are the hypotenuses of the two triangles and are equal, and AM is a side common to both triangles, giving a right angle, equal hypotenuses and one further equal side, exactly RHS, so 'RHS, using AM as common side' is correct. 'SAS, right angle as included angle' wrongly treats the right angle at M as included between AB and AM, but AB is the hypotenuse, not one of the two sides forming that right angle. 'SSS, using BM = CM as a fact' wrongly assumes BM equals CM as a given fact, when this is only true because of the RHS congruence, not before it, so it cannot be used to prove that congruence. 'ASA, AB as the included side' again wrongly labels a side as if it could sit between two angles when only one angle, the right angle, is actually known.
- (c) 22 m — Rearranging P = 2l + 2w for l gives l = (P − 2w)/2. Substituting P = 60 and w = 8: l = (60 − 16)/2 = 44/2 = 22 m. Answering 30 m comes from ignoring the width altogether and working out l = P/2, which only holds when w = 0. Answering 38 m comes from adding 2w instead of subtracting it: (60 + 16)/2 = 38. Answering 11 m comes from dividing by 4 instead of 2, as if the formula were P = 4l + 4w. The pen's length is 22 m.
- (d) 3 : 2 : 5 — The highest common factor of 45, 30 and 75 is 15. Divide each part by 15: 45 ÷ 15 = 3, 30 ÷ 15 = 2 and 75 ÷ 15 = 5, giving 3 : 2 : 5. Giving 9 : 6 : 15 divides by 5, a common factor but not the highest one. Giving 15 : 10 : 25 divides by 3 only, even further from simplest form. Giving 2 : 3 : 5 has the first two parts swapped.
- (a) 225 — Method: find the interior angle of the regular octagon, then subtract it from 360° to find the reflex angle at the same vertex, since the interior angle and the reflex angle together make a full turn. Working: there are 8 − 2 = 6 triangles' worth of angle in the octagon, so the interior angle = 6 × 180 ÷ 8 = 135; reflex angle = 360 − 135 = 225. Answer: 225°. A candidate who stops after finding the interior angle gives 135. A candidate who works out the exterior angle instead, 360 ÷ 8 = 45, gives 45. A candidate who subtracts the exterior angle from 360° instead of the interior angle, working out 360 − 45, gets 315.
- (b) y = −3x/4 + 25 — Method: a straight line that touches a circle at one point is a tangent there, so it is perpendicular to the radius drawn to that point; find the gradient of the radius, take its negative reciprocal, then substitute the point of contact into y − y₁ = m(x − x₁). Working: the radius from (0, 0) to (12, 16) has gradient 16 ÷ 12, which cancels to 4/3, so the tangent has gradient −3/4. Substituting gives y − 16 = −3/4(x − 12), so y − 16 = −3x/4 + 9 and y = −3x/4 + 25. Answer: y = −3x/4 + 25. The distractors: y = 3x/4 + 7 turns the gradient of the radius upside down but leaves it positive, so the perpendicular step is only half done; y = −4x/3 + 32 changes the sign of the radius gradient without turning it upside down, which is the other half left undone; y = −3x/4 − 25 uses the correct gradient but substitutes the point of contact with both signs reversed, writing y + 16 = −3/4(x + 12).
- (b) 10 km — Method: find the constant speed from the graph (distance ÷ time for any point on the line), then multiply that speed by 20 minutes. Working: the line passes through (4 minutes, 2 km), so the speed is 2 ÷ 4 = 0.5 km per minute; in 20 minutes the cyclist travels 0.5 × 20 = 10 km. Answer: 10 km. Distractor refutation: 3 km comes from reading off the distance shown at the end of the plotted section (6 minutes) and stopping there, instead of extending the line to 20 minutes. 20 km comes from misreading the speed as 1 km per minute instead of 0.5 km per minute, doubling the true rate. 40 km comes from dividing 20 by the speed instead of multiplying by it, a reciprocal mix-up.
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.