June 2024 | Edexcel Past Paper I GCSE Maths Higher Paper 2 | Worked Solutions

EDEXCEL GCSE MATHS • HIGHER • PAPER 2H • CALCULATOR

June 2024 Paper 2 — Answers & Worked Solutions

MathsWealth answer guide for Pearson Edexcel GCSE Mathematics Higher Paper 2.

80 marks • 1 hour 30 minutes • Calculator paper

How to use this page: Try each question before reading the written solution. Show the key steps in your own working to help secure method marks. For graphs and constructions, use the original paper grid or diagram.

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Official Exam Materials

Download the original question paper, official mark scheme, and examiner feedback from Pearson Edexcel.

Question Paper (PDF)Mark Scheme (PDF)Examiner Feedback (PDF)

1MA1/2H · Monday 3 June 2024 · 80 marks · 1 hour 30 minutes

Worked Solutions

Topic classification, worked solutions, answers and mark-scheme reminders.

Q1Pythagoras · Point: Find a shorter side of a right-angled triangle.

Question: ABC is a right-angled triangle. The hypotenuse is 19 cm and another side is 10 cm. Find CB, correct to 3 significant figures. (2 marks)

The hypotenuse is 19 cm. By Pythagoras, the shorter side squared is 19² − 10² = 361 − 100 = 261. Take the square root and round at the end.

Answer: 16.2 cm (to 3 significant figures)

Mark scheme focus: M1 for a valid Pythagoras method, e.g. CB² = 19² − 10²; A1 for an answer in the official accepted range 16.1–16.2 cm. Since √261 ≈ 16.1555, the correctly rounded answer to 3 significant figures is 16.2 cm; the official mark scheme also accepts 16.1 cm.

Q2Prime factors and LCM · Point: Write 90 as a product of primes and find an LCM.

Question: (a) Express 90 as a product of prime factors. (b) Given A = 2² × 3 and B = 2 × 3², find the lowest common multiple of A and B. (3 marks)

90 = 2 × 3 × 3 × 5 = 2 × 3² × 5. The given numbers are 2² × 3 and 2 × 3²; use the greatest power of each prime for the LCM: 2² × 3².

Answer: (a) 2 × 3² × 5 (b) 36

Mark scheme focus: (a) M1 for a complete prime-factor method; A1 for 2 × 3² × 5. (b) B1 for LCM = 36.

Q3Inverse proportion · Point: Find the difference in time for two machine counts.

Question: The time H (hours) for n machines to make 5000 bottles is H = 72/n. Six machines work on Monday and nine machines work on Tuesday. Find how much longer the machines take on Monday. (2 marks)

Use H = 72/n. For 6 machines, H = 72 ÷ 6 = 12 hours. For 9 machines, H = 72 ÷ 9 = 8 hours. Difference = 12 − 8.

Answer: 4 hours

Mark scheme focus: M1 for using H = 72/n for both machine counts; A1 for the difference 12 − 8 = 4 hours.

Q4Probability and ratio · Point: Use a probability to find the total number of discs, then use a ratio.

Question: A bag contains red, blue and yellow discs. There are 24 yellow discs and the probability of selecting yellow is 0.16. The ratio red : blue is 5 : 4. Find the number of red discs. (4 marks)

There are 24 yellow discs and P(yellow) = 0.16, so the total number of discs is 24 ÷ 0.16 = 150. Red and blue discs together: 150 − 24 = 126. The ratio red : blue is 5 : 4, so there are 9 parts and each part is 126 ÷ 9 = 14. Red discs = 5 × 14 = 70.

Answer: 70

Mark scheme focus: P1 for finding total discs 24 ÷ 0.16 = 150 (or equivalent probability method); P1 for 126 non-yellow discs; P1 for applying 5:4 to 126; A1 for 70 red discs.

Q5Quadratic graph

Question: (a) Complete the table for y = x² − x at x = −2, −1, 0, 1, 2, 3. (b) Draw the graph for −2 ≤ x ≤ 3. (c) Use the graph to estimate the solutions of x² − x = 4. (6 marks)

Substitute each x-value into y = x² − x:
x = −2: y = 4 + 2 = 6
x = −1: y = 1 + 1 = 2
x = 0: y = 0
x = 1: y = 0
x = 2: y = 2
x = 3: y = 6
Plot (−2, 6), (−1, 2), (0, 0), (1, 0), (2, 2), (3, 6), then draw a smooth curve. For part (c), read the x-coordinates where the curve meets y = 4; they are approximately −1.56 and 2.56 (normally read from the graph as about −1.6 and 2.6).

Answer: Table y-values: 6, 2, 0, 0, 2, 6. Graph: smooth parabola through the six points. (c) x ≈ −1.6 and x ≈ 2.6.

Mark scheme focus: (a) B2 for all missing table values correct (B1 for one or two). (b) B2 for the correct smooth graph / B1 for at least five correct points. (c) M1 for reading intersections with y = 4; A1 for estimates in the ranges −1.7 to −1.5 and 2.5 to 2.7.

Q6Ratios and percentages · Point: Check whether three people finish with equal numbers of sweets after two transfers.

Question: Andy, Luke and Tina share sweets in the ratio 1 : 6 : 14. Tina gives 3/7 of her sweets to Andy, then gives 12.5% of what remains to Luke. Decide whether all three then have the same number, showing the amounts. (4 marks)

Start with the ratio Andy : Luke : Tina = 1 : 6 : 14. Tina gives Andy 3/7 of her 14 parts: 14 ÷ 7 × 3 = 6 parts. The amounts become A = 1 + 6 = 7, L = 6, T = 14 − 6 = 8. Tina then gives Luke 12.5% of her remaining 8 parts: 0.125 × 8 = 1 part. Final amounts: Andy = 7, Luke = 6 + 1 = 7, Tina = 8 − 1 = 7.

Answer: Yes. Each person finishes with 7 equal parts.

Mark scheme focus: P1 for finding Tina gives Andy 6 parts; P1 for tracking remaining sweets; P1 for finding the 12.5% transfer; C1 for “Yes” supported by final amounts 7 : 7 : 7.

Q7Angles and parallel lines · Point: Find the value of x and use the angle result to justify that the lines are parallel.

Question: A quadrilateral ABCD has interior angles (in order) 2x + 15°, 4x + 15°, 4x + 8° and 3x − 3°. Show that ABCD is a trapezium. (4 marks)

Use the angles shown in the diagram to form a valid equation. One accepted route is 4x + 15 + 2x + 15 + 4x + 8 + 3x − 3 = 360, which simplifies to 13x + 35 = 360. Hence 13x = 325 and x = 25. Substitute x = 25. The angles are 65°, 115°, 108° and 72°. In particular, 65° + 115° = 180° (and 108° + 72° = 180°). The co-interior angles on each side are supplementary, so the relevant opposite sides are parallel and ABCD is a trapezium.

Answer: x = 25; ABCD is a trapezium because a pair of co-interior angles is supplementary (65° + 115° = 180°).

Mark scheme focus: M1 for a suitable angle-sum equation; dependent M1 for solving for x; A1 for x = 25; C1 for substituting and showing a relevant pair of co-interior angles sum to 180° with a parallel-lines conclusion.

Q8Scale diagrams · Point: Use a scale drawing and triangle area to find a real length.

Question: A playground is a right-angled triangle. On a scale drawing, 1 cm represents 5 m and the drawing area is 28 cm². The real length QR is 40 m. Find the real length PQ. (3 marks)

The scale is 1 cm to 5 m, so a real length of 40 m is 8 cm on the drawing. The drawing area is 28 cm²: ½ × PQ × 8 = 28, so PQ = 7 cm. Convert back to real length: 7 × 5.

Answer: 35 m

Mark scheme focus: P1 for converting the 40 m real length to 8 cm on the drawing or setting up triangle area correctly; A1 cao for 35 m.

Q9Bounds · Point: Find a lower bound and assess a rounding claim.

Question: N rounds to 7.3 when rounded to 2 significant figures. (a) Give the least possible value of N. (b) Assess the claim that N cannot exceed 7.349 because 7.350 rounds to 7.4. Give a reason. (2 marks)

Since 7.3 is given to 2 significant figures, the rounding interval is 7.25 ≤ N < 7.35. The lower bound 7.25 is included. The upper bound 7.35 is excluded, but values such as 7.3491 are still below 7.35 and therefore round to 7.3. So the claim that N cannot exceed 7.349 is false.

Answer: (a) 7.25 (b) No: values between 7.349 and 7.350 exist / 7.3491 is greater

Mark scheme focus: (a) B1 for 7.25. (b) C1 for “No” with a valid reason, e.g. 7.3491 is greater than 7.349 and still rounds to 7.3, or values exist between 7.349 and 7.350.

Q10Trigonometry and algebra · Point: Solve for x using two trigonometric ratios.

Question: Two right-angled triangles have side lengths 8 and 7 − 2x in one triangle, and 4 and 1 + x in the other. Given sin a = tan b, use the diagram’s ratios to find x. (3 marks)

From the diagram, sin a = (7 − 2x)/8 and tan b = (1 + x)/4. Since sin a = tan b, (7 − 2x)/8 = (1 + x)/4. Multiply by 8: 7 − 2x = 2 + 2x, so 5 = 4x and x = 5/4 = 1.25.

Answer: x = 1.25

Mark scheme focus: B1 for identifying sin a = (7 − 2x)/8 or tan b = (1 + x)/4; P1 for equating and solving the ratios; A1 for x = 1.25.

Q11Cumulative frequency · Point: Complete a cumulative frequency graph and estimate statistics.

Question: The grouped frequency table for the weights of 60 parcels has frequencies 7, 21, 15, 11 and 6 for the intervals 0 < w ≤ 2, 2 < w ≤ 4, 4 < w ≤ 6, 6 < w ≤ 8 and 8 < w ≤ 10 kg. (a) Complete the cumulative frequency table. (b) Draw the cumulative frequency graph. (c) Estimate the interquartile range. (d) Estimate how many parcels weigh more than 7.4 kg. (7 marks)

Add frequencies cumulatively to get 7, 28, 43, 54, 60. Plot cumulative frequency against the upper class boundary, join with a smooth increasing curve, then read quartiles from the graph.

Answer: (a) 7, 28, 43, 54, 60 (c) IQR approximately 3.6 kg (d) approximately 9 parcels

Mark scheme focus: (a) B1 for cumulative frequencies 7, 28, 43, 54, 60. (b) B2 for a correct cumulative frequency graph / B1 for correct plotting. (c) M1 for a quartile reading; A1 for IQR 3.4–4 kg. (d) M1 for reading CF at 7.4 kg; A1 for 7–10 parcels.

Q12Density and direct proportion · Point: Find the constant of proportionality in f = k/d², then calculate d when f = 10.

Question: f is inversely proportional to d². When d = 8, f = 3.5. (a) Find f in terms of d. (b) Find the positive value of d when f = 10, to 3 significant figures. (4 marks)

For part (a), use f = k/d² and substitute f = 3.5 and d = 8: 3.5 = k/8², so k = 3.5 × 64 = 224. Therefore f = 224/d². For part (b), substitute f = 10: 10 = 224/d², so d² = 224 ÷ 10 = 22.4. Taking the positive square root gives d = √22.4 ≈ 4.733, which is 4.73 to 3 significant figures.

Answer: (a) k = 224, so f = 224/d². (b) d ≈ 4.73.

Mark scheme focus: (a) M1 for k = 3.5 × 8² = 224; A1 for f = 224/d². (b) M1 for d² = 224/10; A1 for d = 4.73–4.733 to 3 significant figures.

Q13Inequalities · Point: Shade the region satisfying four inequalities.

Question: On the coordinate grid, shade and label region R satisfying all four inequalities: x ≤ 2, y ≥ −3, y < 2x + 1, and 3x + 2y ≤ 6. (3 marks)

Draw y = −3, x = 2, y = 2x + 1 and 3x + 2y = 6. Draw y = 2x + 1 as a dashed line because y < 2x + 1 is strict; draw the other three boundaries as solid lines. Shade above y = −3, below y = 2x + 1, to the left of x = 2, and on the side of 3x + 2y = 6 where the expression is at most 6. Both vertices on the dashed line are excluded.

Answer: Shade and label the region R. Its corner coordinates are (−2, −3), (2, −3), (2, 0) and (4/7, 15/7). The boundary y = 2x + 1 is dashed, so both (−2, −3) and (4/7, 15/7), which lie on this boundary, are excluded. The other three boundaries are solid and included.

Mark scheme focus: M1 for correct boundary lines; M1 for correct shading; C1 for the correct region R, with y = 2x + 1 dashed and both boundary intersections (−2, −3) and (4/7, 15/7) excluded.

Q14Rates of change · Point: Estimate a gradient from a tangent and area under a curve.

Question: A velocity-time graph shows a car slowing down. (a) Estimate its acceleration at t = 5 seconds using a tangent. (b) Estimate the distance travelled during the first 6 seconds using 3 strips of equal width. (6 marks)

Draw a tangent at t = 5 and calculate its gradient using two well-separated points. For area, split the region into trapezia and use ½(a + b)h for each, then add the areas.

Answer: (a) approximately −2.5 m/s² (accept −2.0 to −2.8 with a tangent drawn) (b) 79 m

Mark scheme focus: (a) M1 for drawing a tangent at t = 5; dependent M1 for a complete gradient method (for example, using a change of 10 over 4 seconds); A1 for a gradient from −2.0 to −2.8, with the tangent drawn. The mark-scheme answer is −2.5 m/s². (b) M1 for a valid trapezium-area estimate, e.g. ½ × 2 × (25 + 16) = 41; or a permitted midpoint-rectangle estimate. A further M1 for a complete method using all three equal-width strips; A1 for 79 m.

Q15Surds · Point: Rationalise the denominator.

Question: Given that a is a prime number, rationalise the denominator of 1/(√a + 1), giving the answer in its simplest form. (2 marks)

Multiply top and bottom by the conjugate of the denominator: 1/(√a + 1) × (√a − 1)/(√a − 1). The denominator becomes a − 1.

Answer: (√a − 1)/(a − 1)

Mark scheme focus: M1 for multiplying numerator and denominator by the conjugate (√a − 1); A1 for (√a − 1)/(a − 1) or an equivalent simplest form.

Q16Inequalities · Point: Solve a compound inequality.

Question: Solve the inequality (4x − 3)(x + 5) < 0. (2 marks)

Solve each boundary equation first. Combine the conditions into one interval, taking care to keep strict inequalities if the original signs are < or >.

Answer: −5 < x < 3/4

Mark scheme focus: M1 for critical values −5 and 3/4; A1 for the interval −5 < x < 3/4.

Q17Ratios of volumes · Point: Compare three similar solids.

Question: L, M and P are similar solid cylinders made from the same material. L has mass 64 g, M has mass 125 g, M has total surface area 144 cm² and P has total surface area 16 cm². Find the ratio height of L : height of M : height of P. (4 marks)

Because the cylinders are made from the same material, mass is proportional to volume. Thus L:M volume = 64:125, so the linear (and height) ratio is ∛64:∛125 = 4:5. For similar cylinders, surface area scales with the square of the linear scale factor. Therefore M:P height ratio = √144:√16 = 12:4 = 3:1. Combine 4:5 and 3:1 using a common M term: L:M = 12:15 and M:P = 15:5. Hence L:M:P = 12:15:5.

Answer: 12 : 15 : 5

Mark scheme focus: P1 for the height ratio L:M from cube roots of the masses (4:5); P1 for M:P from square roots of surface areas (3:1); P1 for combining to a common middle term; A1 for 12:15:5.

Q18Probability without replacement · Point: Find the probability of drawing exactly two red counters and one green counter in any order, or three red counters, when three counters are drawn without replacement.

Question: A bag contains 4 red, 3 yellow and 1 green counter. Tony draws 3 counters at random without replacement. Find the probability that after the draws there are more yellow counters than red counters left in the bag. Show your working. (5 marks)

The mark scheme uses the four disjoint outcomes RRG, RGR, GRR and RRR. Their probabilities are 4/8 × 3/7 × 1/6 = 1/28; 4/8 × 1/7 × 3/6 = 1/28; 1/8 × 4/7 × 3/6 = 1/28; and 4/8 × 3/7 × 2/6 = 1/14. Add them: 1/28 + 1/28 + 1/28 + 1/14 = 5/28. The denominators decrease because the counters are not replaced.

Answer: 5/28

Mark scheme focus: P1 for a correct later-draw probability; P1 for a correct product for a two-red/one-green sequence; P1 for the three-red sequence; P1 for summing the valid disjoint cases; A1 for 5/28.

Q19Vectors · Point: Express a vector in terms of a and b and test parallelism.

Question: In quadrilateral OACB, M is the midpoint of OA and N lies on BC with BN : NC = 4 : 5. Given OA = a, OB = b and AC = kb (k a positive integer), (a) express MN in terms of k, a and b in simplest form; (b) decide whether MN is parallel to OB and explain. (5 marks)

Since OC = OA + AC = a + kb, BC = a + kb − b. As BN : NC = 4 : 5, CN = (5/9)CB = (5/9)(b − a − kb). Also MC = OC − OM = a + kb − (1/2)a = (1/2)a + kb. Therefore MN = MC + CN = (1/2)a + kb + (5/9)(b − a − kb) = −(1/18)a + (4/9)kb + (5/9)b = (1/18)(8kb + 10b − a). This contains a non-zero a-component, so it is not a multiple of b and is not parallel to OB.

Answer: (a) (1/18)(8k b + 10b − a) (b) No, it contains an a-component

Mark scheme focus: P1 for a correct BC-related vector; P1 for the 4:5 division of BC; P1 for an unsimplified MN vector; A1 for (8kb + 10b − a)/18. (b) C1 for “No” with a reason: MN is not a multiple of b / has a non-zero a-component.

Q20Completing the square · Point: Find the turning point of a quadratic.

Question: The curve C has equation y = 2x² − 12x + 7. Find the coordinates of its turning point. (3 marks)

Factor 2 from the x² and x terms, then complete the square: 2x² − 12x + 7 = 2(x² − 6x) + 7 = 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 11.

Answer: Turning point (3, −11)

Mark scheme focus: P1 for starting the completion of the square; P1 for obtaining y = 2(x − 3)² − 11 or equivalent working; A1 for turning point (3, −11).

Q21Graph transformations · Point: Transform a graph using a translation and reflection.

Question: The graph of y = g(x) is shown. On the same grid, draw the graph of y = g(−x) + 2. (2 marks)

A translation 2 units upwards adds 2 to every y-coordinate. Reflection in the y-axis changes every x-coordinate to its opposite. Apply both changes to the original key points and plot them.

Answer: Correct transformed graph; e.g. (−1,1) → (1,3), (1,−2) → (−1,0), (3,1) → (−3,3)

Mark scheme focus: M1 for reflecting y = g(x) in the y-axis to get y = g(−x); A1 for translating the reflected graph up 2 units.

Q22Circle theorem proof · Point: Prove two tangents from the same external point are equal.

Question: A and B lie on a circle with centre O. MAP and NBP are tangents to the circle. Prove that AP = BP. (4 marks)

OA and OB are radii, and a radius meets a tangent at 90°. Triangles OAP and OBP therefore have equal hypotenuse OP and one equal side OA = OB, with right angles. They are congruent by RHS, so AP = BP.

Answer: AP = BP, with a complete RHS congruence proof

Mark scheme focus: M1 for recognising radii are equal and perpendicular to tangents; M1 for identifying the shared hypotenuse OP; M1 for RHS congruence; A1 for concluding AP = BP.

Revision checklist

• Show enough working to earn method marks.
• Keep full calculator precision until the final rounding step.
• Include units and correct mathematical notation.
• For proofs and explanations, link each statement to a reason.

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