2026 GCSE Maths Higher Practice Paper 1 – Worked Solutions

MathsWealth 2026 GCSE Maths

Higher Tier — Practice Paper 1 Worked Solutions

WORKED SOLUTIONS

Paper: Higher Tier Paper 1

Total: 80 marks

Calculator: Not allowed

Questions: 23

How to use this page: Try each question first, then compare your method with the worked solution. The mark notes show the main steps that earn credit; they are illustrative rather than an official Pearson mark scheme.

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Worked SolutionsYou are here

1 [2 marks]

Common difference = 10 − 6 = 4. M1

The first term is 6, so nth term = 6 + 4(n − 1) = 4n + 2. A1

Answer: 4n + 2

2 [4 marks]

(a)

79 = 2836, and 512 = 1536. M1

28 − 1536 = 1336. A1

(b) The numerator and denominator should both be multiplied: 5×3 = 15 and 6×10 = 60, so the fraction is 15/60 = 1/4. B1

Answers: 13/36 and 1/4

3 [5 marks]

(a)

Garden area = 12×7 = 84 m²; pond area = 4×3 = 12 m². Remaining area = 72 m². M1

Paint = 5×2.5 = 12.5 litres. Coverage = 12.5×8 = 100 m². M1

100 > 72, so there is enough paint. A1

(b) At 5 m² per litre, coverage = 12.5×5 = 62.5 m². This is less than the 72 m² needed, so the gardener does not have enough paint. M1 A1

Answer: No. The gardener is short by 72 − 62.5 = 9.5 m² of coverage.

4 [3 marks]

(a) A but not B is 12. A1

(b) Total = 12+8+15+21+6 = 62. A∪B = 12+8+15 = 35.

P(A∪B) = 35/62. M1 A1

Answer: 12 and 35/62.

5 [4 marks]

(a) 468 km → 500 km and 18 km/L → 20 km/L.

Estimated fuel = 500÷20 = 25 litres. M1 A1

(b) Actual fuel = 468÷18 = 26 litres. Since 25 < 26, the estimate is an underestimate. B1

Answer: 25 litres, underestimate.

6 [4 marks]

(a)

Gradient = (10−1)/(4−(−2)) = 9/6 = 3/2. M1

1 = (3/2)(−2)+c, so c=4. A1

(b) A parallel line has the same gradient and passes through (0,−5).

y = (3/2)x − 5. A1

Answers: y = 3x/2 + 4; y = 3x/2 − 5.

7 [5 marks]

Blue cards = 480×5/8 = 300. M1

Small blue cards = 300×2/5 = 120. M1

Small blue maths cards = 120×3/5 = 72. M1

Percentage = (72/480)×100 = 15%. A1 A1

Answer: 15%

8 [3 marks]

Total mass of 8 = 8×2.75 = 22 kg. M1

Total mass of 3 = 3×2.1 = 6.3 kg. Other 5 = 22−6.3 = 15.7 kg. A1

Mean = 15.7÷5 = 3.14 kg. A1

9 [1 mark]

0.72 = 72% of the original price, so the reduction is 100−72 = 28%.

Answer: 28%

10 [4 marks]

From 2x+y=9, y=9−2x. M1

4x−3(9−2x)=13 → 10x=40 → x=4. M1 A1

y=9−8=1. A1

Answer: x=4, y=1

11 [3 marks]

A point P has coordinates (4, −2).

The point P is translated by the vector

−3 5

to give point Q.

(a) Write down the coordinates of Q.

Add −3 to the x-coordinate and 5 to the y-coordinate: Q = (4−3, −2+5) = (1, 3). M1 A1

Point Q is then rotated 90° clockwise about the origin to give point R.

(b) Write down the coordinates of R.

A 90° clockwise rotation about the origin maps (x, y) to (y, −x). Therefore (1, 3) maps to (3, −1). A1

Answers: Q = (1, 3); R = (3, −1).

12 [3 marks]

y=x²−4: opens upwards. A1

y=−x³: decreasing. A1

y=6/x: has no value at x=0. A1

13 [5 marks]

Frequency density = frequency ÷ class width.

Interval	Density
0–10	1.8
10–25	3.6
25–35	3.0
35–50	2.4
50–60	1.2

(b) From 20–25: 5×3.6=18 students estimated. From 25–35: 30. From 35–40: 5×2.4=12.

Estimated number = 18+30+12 = 60. M1 A1

Fraction = 60/150 = 2/5. A1

Answer: 2/5. Histogram bars should have the five densities above. 2 marks

14 [3 marks]

(x+4)(x−3)=x²+x−12. M1

(2x−1)(x²+x−12) = 2x³+x²−25x+12. A1 A1

15 [4 marks]

Arc length rθ = 4π, so 8θ=4π and θ=π/2. M1

Sector area = 1/2 r²θ = 1/2×64×π/2 = 16π cm². M1 A1 A1

16 [2 marks]

First counter red: n/(n+1). After one red is taken, n−1 red remain out of n counters, so second red: (n−1)/n. M1

Multiply: n/(n+1) × (n−1)/n = (n−1)/(n+1). A1

17 [3 marks]

(a)

3√5×√5√5 = 3√5/5. A1

(b)

√80=4√5 and √20=2√5, so √80−√20 = 2√5. M1 A1

18 [3 marks]

0.24 = 24/99 and 0.15 = 15/99. M1

Therefore 0.24 + 0.15 = (24+15)/99 = 39/99. M1 A1

Here m=39.

19 [3 marks]

By Pythagoras in the smaller triangle, the missing side is √(10²−6²)=√64=8 cm. M1

Scale factor = 15/10 = 1.5. M1

Corresponding side = 8×1.5 = 12 cm. A1

20 [3 marks]

From 2^y=3²×2, we have 3²=2^y−1. M1

Then 2ⁿ=2^x×2^y−1=2^x+y−1. M1

Since x+y=7, n=7−1=6. A1

21 [4 marks]

Let width=w and height=h. Volume: 10wh=240, so wh=24. M1

Surface area: 2(10w+10h+wh)=316, so 10(w+h)+24=158 and w+h=13. M1

The two numbers are 4 and 9 because they multiply to 24 and add to 13. Width>height, so h=4. A1

Answer: 4 cm. A1

22 [4 marks]

(a) The cosine graph passes through (0,1), (90,0), (180,−1), (270,0) and (360,1), with a smooth curve. 2 marks

(b) 2 cos x° = −1, so cos x° = −1/2. In 0° ≤ x ≤ 360°, this occurs at 120° and 240°. M1 A1

Answer: x = 120°, 240°.

23 [5 marks]

Circle C has centre (0, 0) and passes through P(6, 8). A line through P meets the circle again at Q(−8, −6).

(a) Work out the radius of circle C.

The centre is (0,0), so use Pythagoras: r = √(6² + 8²) = √100 = 10. M1 A1

Answer (a): 10 units

(b) The line PQ crosses the y-axis at (0, k). Find the value of k. Show all your working.

Gradient of PQ = (8−(−6)) ÷ (6−(−8)) = 14 ÷ 14 = 1. M1

So the line has equation y = x + c. Using P(6,8): 8 = 6 + c, so c = 2. M1 A1

Answer (b): k = 2

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END OF WORKED SOLUTIONS

MathsWealth Original Resource
This is an independently created educational resource. It is not an official Pearson Edexcel examination paper, mark scheme or Pearson publication, and Pearson has not endorsed it. This is an independently created MathsWealth practice resource and is not an official Pearson Edexcel paper or mark scheme.