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%.
GCSE Higher sample Paper 1 (non-calculator)
- 1.A recipe uses 160 g of flour. Sam wants to increase the amount by 1/4. Work out the new amount of flour.
- 2.A colony of bacteria doubles in number every hour. At 9am there are 5 bacteria in the colony. Work out how many bacteria there will be at 12 noon.
- 3.Two numbers are in the ratio 3 : 7. Their sum is 60. Work out the positive difference between the two numbers.
- 4.A(1, 2), B(5, 2) and C(5, 6) are three of the four vertices of a square ABCD. Work out the coordinates of D.
- 5.In a survey of 200 households, 120 have a garden and 80 own a dog. 54 of the households have a garden and own a dog. Work out the probability that a household owns a dog given that it has a garden, and compare it with the probability that a household picked from the whole survey owns a dog.
- 6.A scatter graph of the number of ice creams sold at a seaside kiosk in Bournemouth and the number of sunburn cases treated at a nearby pharmacy, recorded on the same 30 days, shows strong positive correlation. Which statement about this correlation is correct?
- 7.Work out 1 1/2 ÷ 3/4 exactly, giving your answer in its simplest form.
- 8.The point (9, 40) lies on the circle x² + y² = 1681, which has centre (0, 0). Work out the equation of the tangent to the circle at (9, 40), giving your answer in the form ax + by = c.
- 9.Two quantities, P and Q, are in inverse proportion. A graph is drawn with P on the vertical axis and Q on the horizontal axis. Write down which description fits the shape of this graph.
- 10.A cone is cut by a flat plane parallel to its circular base, partway up between the base and the apex. What shape is the cross-section?
- 11.At a school fête, a stall invites visitors to spin a fair spinner with 8 equal sections numbered 1 to 8, and separately toss a fair coin. A visitor wins a small prize only if the spinner lands on a multiple of 3 and the coin lands on heads. Work out the probability that a visitor wins a prize, giving your answer as a fraction in its simplest form.
- 12.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.
- 13.A taxi charges a £3 fixed fee plus £2 for each mile travelled. Write an expression, in pounds, for the total cost of a journey of n miles.
- 14.An ice-cream van's daily takings, in £, are modelled by a curve plotted against the average temperature that day, in °C. At a temperature of 22°C, the gradient of the tangent to this curve is 14. What does this gradient tell you about the takings at 22°C?
- 15.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.
- 16.A car passes the 15 km marker on a road and passes the 195 km marker 3 hours later. Work out the gradient of the distance-time graph, that is the rate of change of distance with time.
- 17.An acid solution has a concentration of 20%. The solution contains 40 g of acid, and the rest of the solution is water. Work out the mass of water in the solution.
- 18.Quadrilateral WXYZ has WX parallel to ZY, and WZ = XY (the two non-parallel sides are equal in length). Give a reason why angle W = angle X.
- 19.The point (−4, 3) lies on the circle x² + y² = 25, which has centre (0, 0). Work out the equation of the tangent to the circle at (−4, 3), giving your answer in the form y = mx + c.
- 20.The point (3, 6) lies on the circle x² + y² = 45. The tangent to the circle at (3, 6) crosses the x-axis at the point P. Work out the coordinates of P.
Answer key
- (a) 200 g — One quarter of 160 g is 40 g. Increasing the amount means adding this on: 160 + 40 = 200 g. Finding the increase, 1/4 of 160 = 40 g, but stopping there without adding it to the original amount leaves just 40 g. Using 4/5 instead of 5/4 as the scaling fraction, 160 × 4/5 = 128 g, actually decreases the amount rather than increasing it. Increasing by a half instead of a quarter, 160 + 80 = 240 g, uses the wrong fraction of 160.
- (a) 40 — From 9am to 12 noon is 3 hours, so the population doubles three times: 5 × 2³ = 40. A candidate who counts the elapsed time as 4 hours (an off-by-one counting error) would reach 5 × 2⁴ = 80. A candidate who counts it as only 2 hours would reach 5 × 2² = 20. A candidate who misreads 'doubles' as 'increases by 2' each hour would compute 5 + 3 × 2 = 11.
- (d) 24 — Method: split 60 into 3 + 7 = 10 equal parts, find the value of one part, then use the difference in ratio parts. Working: 60 ÷ 10 = 6, so the numbers are 3 × 6 = 18 and 7 × 6 = 42, and their difference is 42 − 18 = 24. Answer: 24. 4 comes from finding the difference between the ratio numbers, 7 − 3, but forgetting to multiply by the value of one part. 60 comes from adding the two numbers back together instead of subtracting, which just repeats the given sum. 80 comes from dividing 60 by the first ratio number, 3, instead of by the total number of parts, 10, giving a part value of 20 and a difference of 7 × 20 − 3 × 20 = 80.
- (b) (1, 6) — Method: when a square is set square-on to the grid, so that its sides run parallel to the axes, each vertex shares its x-coordinate with one neighbour and its y-coordinate with the other, and the missing vertex then borrows one coordinate from each of the two vertices it is joined to; so the first job is to check from the given points that the sides really do run parallel to the axes. Working: A(1, 2) and B(5, 2) share y = 2, so AB is a horizontal side; B(5, 2) and C(5, 6) share x = 5, so BC is a vertical side, which confirms that this square lies square-on to the axes and that the rule may be used. In square ABCD the vertex D is joined to C and to A. DC must be horizontal like AB, so D takes the y-coordinate of C, which is 6; DA must be vertical like CB, so D takes the x-coordinate of A, which is 1. D is therefore (1, 6), and checking confirms every side is 4 long. Answer: (1, 6). The distractors: (1, 5) comes from lifting the first number out of each of A and C, pairing the x-coordinate of A with the x-coordinate of C; (6, 1) comes from finding the right two numbers but writing them the wrong way round, height before sideways position; (9, 6) comes from stepping a further 4 to the right from C instead of closing the square back to the column A stands in.
- (a) 0.45, different from 0.4 for all the households — Method: work out the probability inside the restricted group of garden owners, then work out the probability across the whole survey, and compare the two. Working: 54 of the 120 households with a garden own a dog, so the conditional probability is 54 divided by 120, which is 0.45. Across the whole survey 80 of the 200 households own a dog, which is 0.4. Since 0.45 is not 0.4, having a garden changes the chance of owning a dog and the two events are not independent. Answer: 0.45, different from 0.4 for all the households. The distractors: 0.27 is 54/200, dividing the households with both by the whole survey instead of by the 120 with a garden; 0.675 is 54/80, the probability that a household has a garden given that it owns a dog, which is the condition and the event the wrong way round; 0.4 is 80/200, the probability of owning a dog with the garden information never used, which is why that route also reports no difference.
- (c) Neither causes the other; sunshine links both. — Both ice cream sales and sunburn cases tend to rise on hot, sunny days, so the amount of sunshine is a third factor linked to both — neither variable causes the other. Saying ice cream sales cause the sunburn assumes a causal link in one direction that the correlation alone cannot establish. Saying sunburn cases cause the ice cream sales assumes the reverse causal link, which is no more justified. Saying a strong correlation always means causation is the general error this question is testing: correlation, however strong, does not by itself prove that one variable causes the other.
- (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.
- (b) 9x + 40y = 1681 — For a circle x² + y² = r² centred at the origin, the tangent at a point (a, b) on the circle has equation ax + by = r². Here (a, b) = (9, 40) and r² = 1681, so the tangent is 9x + 40y = 1681. Choosing 40x + 9y = 1681 swaps the coefficients, using the y-coordinate as the x-coefficient and the x-coordinate as the y-coefficient. Choosing 9x + 40y = 41 uses the radius 41 instead of r² = 1681 as the constant. Choosing 9x − 40y = 1681 has the correct coefficients and constant but the wrong sign on the y-term.
- (d) A falling curve that never touches either axis — Method: inverse proportion means the product of the two quantities is constant, so P = k ÷ Q; as Q grows P shrinks, and P can never reach zero because k divided by a number is never zero. Working: taking k = 12 as an example, the pairs (1, 12), (2, 6), (3, 4), (6, 2) and (12, 1) drop steeply at first and then flatten out, so the graph is a curve that approaches both axes without meeting either of them. Answer: a falling curve that never touches either axis. The distractors: 'a straight line through the origin' is the graph of direct proportion, P = kQ, which is the opposite relationship; 'a straight line with a negative gradient' is the commonest error, reading 'P falls as Q rises' as a straight line, but on such a line P would drop by the same amount for every increase in Q and would cross the horizontal axis into negative values; 'a straight line crossing the vertical axis above zero' is a relationship of the form P = mQ + c, in which P and Q are not proportional at all.
- (a) A smaller circle — Since the cutting plane is parallel to the circular base, the cross-section is also a circle, but smaller than the base because the cone narrows as it rises towards the apex, so 'a smaller circle' is correct. 'A triangle' wrongly describes the outline seen from the side of the cone, not a horizontal cross-section. 'An ellipse' would only result from a cut made at an angle to the base, not one parallel to it. 'The same size circle as the base' wrongly ignores that the cone tapers, so any parallel cross-section above the base must be smaller.
- (d) 1/8 — The multiples of 3 from 1 to 8 are 3 and 6, so the probability of that event is 2/8, which simplifies to 1/4. The probability of the coin landing on heads is 1/2. Since the spin and the toss are independent, multiply the two probabilities: 1/4 × 1/2 = 1/8. A candidate who answers 1/4 has considered only the spinner and forgotten to combine it with the coin toss. A candidate who answers 1/2 has considered only the coin and forgotten the spinner condition entirely. A candidate who answers 1/16 has counted only one number, 6, as a multiple of 3 instead of two, giving 1/8 × 1/2.
- (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.
- (b) 2n + 3 — Method: multiply the cost per mile by the number of miles to get an expression, then add the fixed fee as a separate term. Working: n miles at £2 each is 2n; add the £3 fixed fee: 2n + 3. Answer: 2n + 3. 3n + 2 comes from swapping the fee and the rate round, treating £3 as the rate per mile and £2 as the fixed fee. 5n comes from adding the fee and the rate together first (3 + 2 = 5) and multiplying the result by n, instead of keeping the fixed fee as its own term. 2n − 3 comes from subtracting the fixed fee instead of adding it.
- (b) Takings rise about £14 per 1°C rise — The gradient here is positive, so as temperature rises, takings rise too: near 22°C, takings increase by about £14 for every 1°C rise in temperature. Reversing this to say takings rise for every 1°C FALL gets the direction of the independent variable backwards — a positive gradient means both quantities move the same way. Saying 'takings are £14 at 22°C' confuses the gradient, a rate of change, with the y-value on the graph, which is the takings itself. Saying takings 'rose £14 in total' from 0°C to 22°C treats the gradient at a single point as if it applied over the whole range from 0°C to 22°C, when it only describes the instant at 22°C. Always keep a rate, a total change and a single reading separate.
- (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.
- (b) 60 km/h — Method: the gradient of a distance-time graph is the change in distance divided by the change in time, and for a journey at a steady rate that gradient is the speed. Working: the change in distance is 195 − 15 = 180 km and the change in time is 3 hours, so the gradient is 180 ÷ 3 = 60 km/h. Answer: 60 km/h. The distractors: 180 km/h comes from stopping at the change in distance and never dividing by the 3 hours; 195 km/h comes from reading the final marker as the rate instead of working with the change between the two markers; 3 km/h comes from quoting the time taken, which belongs on the bottom of the fraction, as though it were the value of the fraction itself.
- (a) 160 g — Method: use the ratio 20:100 to find the mass of the whole solution from the mass of acid, then take the acid away to leave the water. Working: 20:100 = 40:m, and 40 ÷ 20 = 2, so m = 2 × 100 = 200 g of solution; the water is 200 − 40 = 160 g. Answer: 160 g. The distractors: 200 g is the mass of the whole solution, which is the middle step and includes the acid the question asks you to leave out; 8 g comes from working out 20% of 40 g, which treats the 40 g as the whole solution rather than as the 20% inside it; 10 g comes from reading the 40 g as the 80% that is water, giving a solution of 50 g and a difference of 50 − 40.
- (a) Isosceles trapezium: base angles are equal — WX is parallel to ZY and the two non-parallel sides WZ and XY are equal in length, so WXYZ is an isosceles trapezium. In an isosceles trapezium the two angles at each of the parallel sides are equal, so angle W = angle X (and angle Z = angle Y). A student who answers with the parallelogram property has quoted a fact that is true of a parallelogram, but WZ and XY are given as non-parallel so WXYZ is not a parallelogram — and in a parallelogram "opposite angles" pairs W with Y, not W with X. A student who answers with the kite property has again taken a true fact about the wrong shape: a kite's equal sides are two pairs of adjacent sides, not the two non-parallel sides of a trapezium. A student who answers with the rhombus property has used something no rhombus has — a rhombus has two pairs of parallel sides and only two pairs of equal angles; all four angles are equal only in a square.
- (a) y = (4/3)x + 25/3 — The radius from (0, 0) to (−4, 3) has gradient 3 ÷ (−4) = −3/4. The tangent is perpendicular to the radius, so its gradient is the negative reciprocal, 4/3. Using y − y₁ = m(x − x₁) with the point (−4, 3): y − 3 = (4/3)(x + 4), so y = (4/3)x + 16/3 + 3 = (4/3)x + 25/3. Using the radius's own gradient, −3/4, instead of taking the perpendicular gradient, gives y − 3 = (−3/4)(x + 4), which simplifies to y = −(3/4)x once the −3 and +3 in the constant cancel out. Taking the reciprocal of the radius's gradient but keeping the wrong sign, using −4/3 instead of 4/3, gives y = −(4/3)x − 7/3. Correctly finding the gradient 4/3 and expanding the bracket, but forgetting to add the y-coordinate 3 at the end, gives y = (4/3)x + 16/3.
- (a) (15, 0) — Method: the tangent is perpendicular to the radius at the point of contact, so find the gradient of the radius, take its negative reciprocal, write the equation of the tangent, then substitute y = 0 because every point on the x-axis has y-coordinate 0. Working: the radius from (0, 0) to (3, 6) has gradient 6 ÷ 3 = 2, so the tangent has gradient −1/2. Substituting into y − 6 = −1/2(x − 3) gives y = −0.5x + 7.5. Setting y = 0 gives 0.5x = 7.5, so x = 15 and P is (15, 0). Answer: (15, 0). The distractors: (0, 7.5) is where the same tangent crosses the y-axis, reached by setting x = 0 instead of y = 0; (0, 0) comes from using the gradient of the radius, 2, for the tangent, which gives the line y = 2x through the centre and so crosses the x-axis at the origin; (6, 0) comes from changing the sign of the radius gradient without turning it upside down, which gives y = −2x + 12.
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.