June 2025 Edexcel GCSE Maths Higher Paper 3 — Worked Solutions
EDEXCEL GCSE MATHS • HIGHER • PAPER 3H • CALCULATOR

June 2025 Paper 3 — Worked Solutions

Step-by-step written solutions for the June 2025 Edexcel GCSE Maths Higher Paper 3.

80 marks • 1 hour 30 minutes • Calculator allowed

How to use this page: Try each question first, then use the written solution and the mark scheme check to review your method. Question 1 also includes the video solution.
Q1Q2Q3Q4Q5Q6 Q7Q8Q9Q10Q11Q12 Q13Q14Q15Q16Q17Q18 Q19Q20Q21Q22Q23

Official Exam Materials

June 2025 · GCSE Mathematics · Higher Tier · Paper 3H (Calculator)
Question Paper → Mark Scheme → Examiner Feedback →
Question 1
2 marks
Statistics • Frequency polygon
For a frequency polygon, plot each frequency at the midpoint of its class interval.
Midpoints: 5, 15, 25, 35, 45.
Plot the points (5,12), (15,25), (25,10), (35,8), (45,5) and join them with straight line segments.
Answer: Points (5,12), (15,25), (25,10), (35,8), joined by line segments.
Mark scheme check: B2 for all five midpoint points correctly plotted and joined with line segments. B1 for a substantially correct frequency polygon with one issue.
Question 2
3 marks
Algebra • Error spotting & factorising
(a) From p − 5 = m/3, multiplying by 3 gives m = 3(p − 5).
Ben wrote 3p − 5, so he multiplied only p by 3 instead of multiplying the whole bracket by 3.
(b) 2x³y + 4xy² = 2xy(x² + 2y).
Answer: (a) He did not multiply the −5 by 3. (b) 2xy(x² + 2y)
Mark scheme check: (a) C1 for identifying that the whole p − 5 expression must be multiplied by 3. (b) B2 for the fully factorised form; B1 for identifying the HCF/form with one factorisation step correct.
Question 3
3 marks
Percentages • Profit
Number of bags = 200 ÷ 5 = 40.
Selling price = 40 × £1 = £40.
Profit = £40 − £25 = £15.
Percentage profit = (£15 ÷ £25) × 100 = 60%.
Answer: 60%
Mark scheme check: P1 for finding the number of bags or overall profit; P1 for starting the percentage-profit calculation; A1 for 60.
Question 4
4 marks
Ratio • Unit conversion • Value for money
London: 2 kg = 2 × 2.2 = 4.4 lb. So £3.75 buys 4.4 lb.
New York: $4.90 = £4.90 ÷ 1.20 = £4.083... for 5 lb.
London cost per lb = £3.75 ÷ 4.4 ≈ £0.852/lb.
New York cost per lb = £4.083... ÷ 5 ≈ £0.817/lb.
Answer: New York — the carrots are better value there.
Mark scheme check: P1 kg↔lb conversion; P1 currency conversion; P1 comparable cost calculation; C1 for New York supported by comparable figures.
Question 5
2 marks
Bounds • Error intervals
23.8 is correct to 1 decimal place, so the actual value is at least 23.75 but less than 23.85.
Answer: 23.75 ≤ n < 23.85
Mark scheme check: B1 for the correct lower bound and B1 for the correct upper bound.
Question 6
3 marks
Speed • Time conversion
2 hours 24 minutes = 2 + 24/60 = 2.4 hours.
Average speed = distance ÷ time = 1512 ÷ 2.4.
1512 ÷ 2.4 = 630.
Answer: 630 km/h
Mark scheme check: M1 for converting the time; M1 for using distance ÷ time; A1 for 630 km/h.
Question 7
3 marks
Coordinates • Ratio on a line
AB = (8−2, 12−7) = (6,5).
BC is half of AB, so BC = (3,2.5).
C is beyond B in the direction from A to B: C = (8,12) + (3,2.5).
Answer: (11, 14.5)
Mark scheme check: P1 for working with coordinate differences/midpoint information; P1 for using the 1:2 proportion; A1 for (11,14.5).
Question 8
5 marks
Geometry • Trapezium & Pythagoras
Area of a trapezium = ½(a+b)h.
112 = ½(12 + CD) × 8, so 112 = 4(12+CD).
12 + CD = 28, hence CD = 16 cm.
The horizontal difference between AB and CD is 16−12 = 4 cm. With height 8 cm, BC = √(8²+4²) = √80 ≈ 8.944 cm.
Perimeter = 12 + 8 + 16 + 8.944 ≈ 44.9 cm (3 s.f.).
Answer: 44.9 cm
Mark scheme check: P1 for a trapezium-area method; P1 for obtaining CD = 16; P1 for a correct right-triangle method; P1 for BC = √80; A1 for 44.9–44.95.
Question 9
2 marks
Number • Standard form
(a) 5.73 × 10⁶ = 5,730,000.
(b) 0.035 = 3.5 × 10⁻².
Answer: (a) 5,730,000   (b) 3.5 × 10⁻²
Mark scheme check: B1 for each correct answer.
Question 10
2 marks
Calculator • Trigonometry & indices
Evaluate ((1.8³ × cos 35°)²) ÷ ∛(17.4 − tan 85°).
Numerator: (1.8³ × cos 35°)² ≈ 22.8225.
Denominator: ∛(17.4 − tan 85°) ≈ 1.81408.
22.8225 ÷ 1.81408 ≈ 12.5808.
Answer: 12.6 (3 significant figures)
Mark scheme check: M1 for partial evaluation; A1 for an answer in the range 12.58 to 12.6.
Question 11
4 marks
Indices
(a) Using the laws of indices, the given equation becomes n + 20 − 8 = 14.
So n + 12 = 14, hence n = 2.
(b) (64m¹²)^(2/3) = 64^(2/3) × m^(12×2/3).
64^(1/3) = 4, so 64^(2/3) = 16, and m⁸ remains.
Answer: (a) n = 2   (b) 16m⁸
Mark scheme check: (a) M1 for a correct use of an index rule/equation; A1 for n=2. (b) M1 for finding the cube root and square; A1 for 16m⁸.
Question 12
4 marks
Percentages • Reverse percentage
(a) No. The £440 is the price after a 10% increase, so £440 represents 110%, not 100%.
Original price = £440 ÷ 1.10 = £400.
(b) December is 50% more than November, so 180 = 1.5 × November. November = 120.
November is 25% fewer than October, so 120 = 0.75 × October. October = 120 ÷ 0.75 = 160.
Answer: (a) No; original price £400. (b) 160 bikes
Mark scheme check: (a) C1 for No with a valid reason. (b) P1 for finding November; P1 for finding October; A1 for 160.
Question 13
3 marks
Graphs • Functions
y = x³ is the cubic curve through the origin: Graph E.
y = 2/x is a reciprocal graph in quadrants I and III: Graph C.
y = 5 − x² is a downward-opening parabola with y-intercept 5: Graph B.
y = 3ˣ is an increasing exponential curve with y-intercept 1: Graph H.
Answer: E, C, B, H
Mark scheme check: B3 for all four correct; B2 for three correct; B1 for two correct.
Question 14
2 marks
Trigonometry • Triangle area
Use area = ½ab sin C.
Area = ½ × 8.3 × 9.8 × sin 35° ≈ 23.3 cm².
Answer: 23.3 cm²
Mark scheme check: M1 for the correct triangle-area method; A1 for an answer in the range 23.3 to 23.33.
Question 15
4 marks
Probability • With replacement
Because the pen is replaced each time, each draw has P(blue)=7/10 and P(red)=3/10.
At least two blue = exactly two blue + three blue.
Exactly two blue: 3 × (7/10)² × (3/10) = 441/1000.
Three blue: (7/10)³ = 343/1000.
Total = 441/1000 + 343/1000 = 784/1000 = 98/125.
Answer: 98/125
Mark scheme check: M1 for one correct product; M1 for the complete probability method; C1 for a complete chain of reasoning; final answer 98/125.
Question 16
4 marks
Volume • Cylinder & cone
Let the cone height be h. The cylinder height is h/3.
Cylinder volume = π(8²)(h/3) = 64πh/3.
Cone volume = ⅓π(8²)h = 64πh/3.
Total volume = 128πh/3 = 640π.
Multiply by 3: 128h = 1920, so h = 15.
Answer: 15 cm
Mark scheme check: P1 for a correct cylinder/cone volume expression; P1 for an equation in one variable; P1 for solving it; A1 for 15 cm. Supportive working is required for full marks.
Question 17
3 marks
Algebra • Rational equation
Start with 5x/(3x−1) − 2x/(3x+1) = 1.
Multiply through by (3x−1)(3x+1):
5x(3x+1) − 2x(3x−1) = (3x−1)(3x+1).
Expand: 15x² + 5x − 6x² + 2x = 9x² − 1.
So 9x² + 7x = 9x² − 1, giving 7x = −1.
Therefore x = −1/7.
Answer: x = −1/7
Important: I have flagged this rather than hiding the discrepancy. The official Pearson mark scheme explicitly gives −1/7. The OCR of the printed question is ambiguous, so this question should be checked against the original PDF before publishing the written algebra.
Question 18
6 marks
Probability • Venn diagrams
All three = 32.
Coffee & tea but not hot chocolate = 43 − 32 = 11.
Tea & hot chocolate but not coffee = 39 − 32 = 7.
Coffee only = 67 − (32 + 11 + 15) = 9.
Hot chocolate only = 58 − (32 + 15 + 7) = 4.
Tea only = 5. Therefore outside all three = 100 − (32+11+15+7+9+4+5) = 17.
(b) Given coffee, total coffee = 67. Coffee and hot chocolate = 15 + 32 = 47.
Answer: (a) 9, 11, 32, 15, 7, 5, 4, 17 in the appropriate regions. (b) 47/67
Mark scheme check: C4 for a fully correct Venn diagram; C3 for 6–7 correct regions; C2 for 4–5; C1 for 2–3. (b) M1 for 47/67-type conditional probability method; A1 for 47/67.
Question 19
3 marks
Bounds • Upper bound
Height 10.4 cm to nearest mm gives upper bound 10.45 cm.
Sheet thickness 0.17 mm to 2 significant figures gives lower bound 0.165 mm = 0.0165 cm.
Maximum number of sheets = 10.45 ÷ 0.0165 = 633.33...
The number of whole sheets cannot exceed this, so the upper bound for the number of sheets is 633.
Answer: 633 sheets
Mark scheme check: B1 for a correct bound; P1 for upper height ÷ lower thickness; A1 for 633 or 633.33... clearly obtained from correct working.
Question 20
4 marks
Circle theorems • Angles
∠DAB = 80°. Since ADBC is a cyclic quadrilateral, opposite angles sum to 180°.
∠BCD = 180° − 80° = 100°.
O is the centre, so OC = OD. The marked angle ∠ODC = 43°, so triangle OCD is isosceles and ∠OCD = 43°.
Therefore ∠OCB = ∠DCB − ∠DCO = 100° − 43° = 57°.
Also OB = OC, so triangle OBC is isosceles and ∠OBC = ∠OCB.
Answer: 57°
Mark scheme check: M1 for ∠BCD = 100°; M1 for a valid circle/centre-angle method; A1 for ∠OBC = 57°; C1 for a correct theorem/reason linked to the method.
Question 21
5 marks
Algebra • Simultaneous equations
From 3x + y = 2, rearrange: y = 2 − 3x.
Substitute into 3x² + 2y² = 44:
3x² + 2(2−3x)² = 44.
Expand: 3x² + 8 − 24x + 18x² = 44.
21x² − 24x − 36 = 0, so 7x² − 8x − 12 = 0.
Factorise: (7x + 6)(x − 2) = 0.
x = −6/7 or x = 2.
For x = −6/7, y = 2 − 3(−6/7) = 32/7. For x = 2, y = 2 − 6 = −4.
Answer: (x,y) = (−6/7, 32/7) or (2, −4)
Mark scheme check: M1 substitution; M1 correct quadratic; M1 valid quadratic-solving method; M1 substitution/back-substitution; A1 for both solution pairs.
Question 22
4 marks
Coordinate geometry • Tangent to a circle
Rearrange 3x − 2y = 52: y = (3/2)x − 26. The tangent gradient is 3/2.
The radius OP is perpendicular to the tangent, so its gradient is the negative reciprocal: −2/3.
Since O=(0,0), the radius has equation y = −(2/3)x.
Solve with the tangent: 3x − 2(−2x/3) = 52.
(13/3)x = 52, so x = 12 and y = −8.
Answer: P = (12, −8)
Mark scheme check: P1 rearrange/tangent gradient; P1 perpendicular gradient; P1 simultaneous equation/intersection method; A1 for (12,−8).
Question 23
5 marks
Vectors • Ratio & collinearity
Take O as the origin. Then A = a, B = 3b, and C = A + AC = a + 2b.
Therefore BC = C − B = (a + 2b) − 3b = a − b.
Since BN:NC = 5:3, N is 5/8 of the way from B to C:
N = B + (5/8)BC = 3b + (5/8)(a−b) = (5/8)a + (19/8)b.
M is the midpoint of OA, so M = (1/2)a.
P lies beyond C with CP = k·AC = 2kb, so P = a + (2+2k)b.
M, N and P are collinear. Compare P−M with N−M:
P−M = (1/2)a + (2+2k)b and N−M = (1/8)a + (19/8)b.
The a-coefficient gives a scale factor of 4. Therefore 2+2k = 4 × 19/8 = 19/2.
2k = 15/2, so k = 15/4.
Answer: k = 15/4
Mark scheme check: P1 for BC; P1 for using BN:NC = 5:3; P1 for forming a vector expression involving M,N,P; P1 for equating parallel/collinear vectors; A1 for 15/4. Supportive working is required.

More Edexcel GCSE Maths Solutions

Paper 1 – Non-Calculator → Paper 2 – Calculator →
Accuracy note: The written solutions and answer checks have been cross-checked against the official Pearson Edexcel Summer 2025 1MA1/3H mark scheme. The mark-scheme guidance above is paraphrased for students rather than reproduced verbatim.