2026 GCSE Maths Higher Paper 2 – Worked Solutions
MathsWealth · Higher Tier · Calculator · 80 marks
1[2 marks]
Use Pythagoras' theorem.
BC² = 19² − 10² = 361 − 100 = 261
BC = √261 = 16.155...
Answer: 16.2 cm (3 s.f.)
2[3 marks]
(a) 180 = 18 × 10 = 2² × 3² × 5.
Answer: 2² × 3² × 5
(b) A = 2³×3 and B = 2×3². Take the highest power of each prime.
LCM = 2³ × 3² = 72
Answer: 72
3[2 marks]
Inverse proportion gives H = k/n.
9 = k/8 ⇒ k = 72
H = 7212 = 6
Answer: 6 hours
4[4 marks]
30 is 20% of the total number of counters.
Total = 30 ÷ 0.20 = 150
So red + blue = 150 − 30 = 120.
The ratio red : blue = 3 : 2, so there are 5 equal parts.
Red = 120 × 35 = 72
Answer: 72 red counters
5[6 marks]
For y = x² − 2x − 3:
x = −2 → y = 5; −1 → 0; 0 → −3; 1 → −4; 2 → −3; 3 → 0; 4 → 5.
Plot these points and join them with a smooth parabola.
For y = 2, solve x² − 2x − 3 = 2.
x² − 2x − 5 = 0
x = 1 ± √6 ≈ −1.45, 3.45
x = 1 ± √6 ≈ −1.45, 3.45
Answers: table values 5, 0, −3, −4, −3, 0, 5; x ≈ −1.45 and 3.45
6[4 marks]
Total ratio parts = 14, so each part is 420 ÷ 14 = 30.
Ali = 60, Ben = 150, Chloe = 210
Chloe gives 37 of 210 = 90 to Ali, leaving 120. She then gives 20% of 120 = 24 to Ben.
Chloe = 96, Ben = 174
Answer: No. Chloe has 96 sweets and Ben has 174.
7[4 marks]
The angles in a quadrilateral add to 360°.
(3x + 10) + (2x + 20) + (3x + 20) + (2x + 30) = 360
10x + 80 = 360 ⇒ x = 28
Therefore ∠A = 3(28) + 10 = 94° and ∠D = 2(28) + 30 = 86°.
∠A + ∠D = 94° + 86° = 180°
These are co-interior angles on transversal AD. Therefore AB is parallel to CD.
Answer: x = 28 and AB ∥ CD
8[3 marks]
The scale is 1 cm : 4 m. The real 24 m side is therefore 6 cm on the drawing.
Area = ½ × 6 × x = 18
3x = 18 ⇒ x = 6 cm
Real length = 6 × 4 = 24 m.
Answer: 24 m
9[2 marks]
To round to 5.6 to 2 significant figures, the range is 5.55 ≤ N < 5.65.
Answer: least possible value = 5.55
5.65 rounds to 5.7, not 5.6.
5.65 is excluded because the upper boundary is not included.
10[3 marks]
sin a = 711.
Given sin a = tan b, tan b = 711. With opposite side x+2 and adjacent side 6:
(x + 2)/6 = 711
11x + 22 = 42 ⇒ x = 2011 ≈ 1.82
Answer: x ≈ 1.82
11[7 marks]
Cumulative frequencies are 8, 24, 42, 54, 60.
Answer: 8, 24, 42, 54, 60
Plot (2,8), (4,24), (6,42), (8,54), (10,60) and draw a smooth cumulative-frequency curve.
Quartiles occur at CF = 15 and CF = 45. Read the corresponding weights from the graph to estimate Q1 and Q3; the interquartile range is Q3 − Q1.
For 7.2 kg, read the cumulative frequency from the graph and subtract it from 60 to estimate the number above 7.2 kg.
Answers for (c) and (d): read from the completed graph.
12[4 marks]
f is inversely proportional to d², so f = k/d².
5 = k/36 ⇒ k = 180
Equation: f = 180/d²
When f = 12:
12 = 180/d² ⇒ d² = 15 ⇒ d = √15 ≈ 3.87
Answer: d ≈ 3.87 (3 s.f.)
13[3 marks]
Draw the boundary lines x = 3, y = −2, y = 2x + 1 and 2x + y = 8. Because all inequalities are ≤ or ≥, use solid boundary lines.
Test a point such as (0,0) against the inequalities and shade the common region satisfying all four conditions.
Answer: the common feasible region satisfying all four inequalities.
14[6 marks]
From the graph, around t = 4 s the velocity is about 41.2 m/s at t = 2 and 22.5 m/s at t = 6.
Acceleration ≈ (22.5 − 41.2) ÷ (6 − 2) ≈ −4.67 m/s²
Answer: approximately −4.7 m/s²
For the first 6 seconds, use three trapezia of width 2 s.
Distance ≈ 2[(46.1 + 41.2)/2 + (41.2 + 33.3)/2 + (33.3 + 22.5)/2]
≈ 217.6 m
Answer: approximately 218 m
15[2 marks]
1/(√a + 1) × (√a − 1)/(√a − 1)
= (√a − 1)/(a − 1)
Answer: (√a − 1)/(a − 1)
16[2 marks]
(3x − 2)(x + 4) < 0
The critical values are x = 23 and x = −4. The product is negative between the roots.
Answer: −4 < x < 23
17[4 marks]
The cylinders are similar and made from the same material, so mass is proportional to volume.
54 : 128 = 27 : 64 = 3³ : 4³
Therefore the linear scale factor A : B is 3 : 4.
Surface area is proportional to the square of the linear scale factor.
B : C surface area = 160 : 90 = 16 : 9
So the linear scale factor B : C is 4 : 3.
A : B : C = 3 : 4 : 3
Answer: 3 : 4 : 3
18[5 marks]
Initially there are 5 red, 4 yellow and 3 green counters, so there are 12 counters in total.
After three counters are removed, we need the number of yellow counters remaining to be greater than the number of red counters remaining.
4 − y > 5 − r ⇒ r > y + 1
With three counters removed, the only possible favourable colour combinations are RRG and RRR.
Favourable selections = C(5,2) × C(3,1) + C(5,3) = 10×3 + 10 = 40
Total selections = C(12,3) = 220
P = 40220 = 211
Answer: 211
19[5 marks]
Let O be the origin, OA = a and OB = b. Since M is the midpoint of OA, OM = ½a.
N divides AB in the ratio AN:NB = 2:3, so AN = 25 AB.
ON = a + 25(b − a) = 3/5a + 2/5b
MN = ON − OM = 1/10a + 2/5b
Answer: MN = 110 a + 25 b
For MN to be parallel to OB, its a-component would have to be zero. It is not, so MN is not parallel to OB.
20[3 marks]
y = 2x² − 12x + 7 = 2(x² − 6x) + 7
= 2(x − 3)² − 18 + 7 = 2(x − 3)² − 11
Answer: turning point = (3, −11)
21[2 marks]
For y = g(−x), reflect the graph of g in the y-axis. Then +2 moves the graph 2 units upwards.
Answer: reflect in the y-axis, then translate upwards by 2 units.
22[4 marks]
PA and PB are tangents from the same external point P. Radii OA and OB are perpendicular to the tangents.
Triangles OAP and OBP are right-angled. OP is common and OA = OB because both are radii.
By RHS congruence, triangle OAP ≅ triangle OBP.
Answer: PA = PB
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