GCE 2026 Mathematics Paper 2: Worked Solutions

Use the product of the two denominators as the common denominator:

$$\begin{aligned} &\frac{7}{x-3} - \frac{8}{1-2x} \\ &= \frac{7(1-2x) - 8(x-3)}{(x-3)(1-2x)} \end{aligned}$$

Expand the numerator carefully. The minus sign multiplies both terms of $8(x-3)$:

$$= \frac{7 - 14x - 8x + 24}{(x-3)(1-2x)} = \frac{31 - 22x}{(x-3)(1-2x)}$$

Answer: $\dfrac{31 - 22x}{(x-3)(1-2x)}$

(i) After the first ball is taken, only 4 balls remain, so every second-ball probability has denominator 4.

(ii) “Different colours” means green then blue or blue then green:

$$\begin{aligned} P(\text{different}) &= P(GB) + P(BG) \\ &= \frac{2}{5}\times\frac{3}{4} + \frac{3}{5}\times\frac{2}{4} \\ &= \frac{6}{20} + \frac{6}{20} = \frac{12}{20} \end{aligned}$$

Answer: $\dfrac{3}{5}$

Always fill a three-set Venn diagram from the centre outwards.

• All three: $4$
• Mangoes and bananas only: $9 - 4 = 5$; mangoes and guavas only: $10 - 4 = 6$; bananas and guavas only: $7 - 4 = 3$
• Mangoes only: $25 - (5 + 4 + 6) = 10$; bananas only: $20 - (5 + 4 + 3) = 8$; guavas only: $15 - (6 + 4 + 3) = 2$
• None: $40 - (10 + 8 + 2 + 5 + 6 + 3 + 4) = 40 - 38 = 2$

(ii) (a) Guavas only: 2
(b) Exactly two types: $5 + 6 + 3 =$ 14
(c) Do not like bananas: $40 - 20 =$ 20

(i) For $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ the determinant is $ad - bc$:

$$\begin{aligned} 3(x-3) - (-1)(x+2) &= 5 \\ 3x - 9 + x + 2 &= 5 \\ 4x &= 12 \end{aligned}$$

Answer: $x = 3$

(ii) When $x = 3$, $\mathbf{M} = \begin{pmatrix} 0 & -1 \\ 5 & 3 \end{pmatrix}$. Swap the leading diagonal, change the signs of the other two entries and divide by the determinant:

$$\mathbf{M}^{-1} = \frac{1}{5}\begin{pmatrix} 3 & 1 \\ -5 & 0 \end{pmatrix} = \begin{pmatrix} \tfrac{3}{5} & \tfrac{1}{5} \\ -1 & 0 \end{pmatrix}$$

(i) Using $T_n = ar^{n-1}$: $\;ar = 6 \;\ldots(1)\;$ and $\;ar^3 = 1.5 \;\ldots(2)$

Divide (2) by (1) to remove $a$:

$$\frac{ar^3}{ar} = \frac{1.5}{6} \;\Rightarrow\; r^2 = \frac{1}{4} \;\Rightarrow\; r = \pm\frac{1}{2}$$

With $r = 0.5$: $a = \dfrac{6}{0.5} = 12$.

Answer: $r = 0.5$ and $a = 12$ (or $r = -0.5$ and $a = -12$)

(ii) $T_8 = ar^7 = 12 \times (0.5)^7 = 0.09375$. Answer: $0.0938$ (3 s.f.)

(iii) $S_\infty = \dfrac{a}{1-r} = \dfrac{12}{1-0.5} = 24$. Answer: $24$ (with $r = -0.5$, $a = -12$: $S_\infty = -8$)

Increase each power by 1 and divide by the new power:

$$\begin{aligned} &\int_{-1}^{1} (3 - 4x + 9x^2 - 4x^3)\,dx \\ &= \Big[\,3x - 2x^2 + 3x^3 - x^4\,\Big]_{-1}^{1} \end{aligned}$$ $$\begin{aligned} &= (3 - 2 + 3 - 1) - (-3 - 2 - 3 - 1) \\ &= 3 - (-9) = 12 \end{aligned}$$

Answer: $12$

• (a)(ii) PR = 7.7 cm. You can check this with the cosine rule: $PR^2 = 10^2 + 7^2 - 2(10)(7)\cos 50^\circ = 59.0$, so $PR \approx 7.68$ cm.
• (b)(i) An arc of radius 6 cm with centre Q.
• (b)(ii) A line parallel to QR, 2 cm from it, inside the triangle.
• (b)(iii) Points equidistant from two lines lie on the bisector of the angle between them, here the bisector of angle QPR.
• (c) T lies inside the arc, on the far side of the parallel line from QR, and on the PR side of the angle bisector: the shaded region.

Find a route along known vectors from the start point to the end point.

• (i) $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = 2\mathbf{a} + 2\mathbf{b}$
• (ii) $\overrightarrow{MA} = \mathbf{a}$ (half of OA), so $\overrightarrow{MB} = \overrightarrow{MA} + \overrightarrow{AB} = \mathbf{a} + 2\mathbf{b}$
• (iii) AN = 2NB means N is $\frac{2}{3}$ of the way from A to B: $\overrightarrow{AN} = \frac{2}{3} \times 2\mathbf{b} = \frac{4}{3}\mathbf{b}$, so $\overrightarrow{ON} = 2\mathbf{a} + \frac{4}{3}\mathbf{b}$
• (iv) $\overrightarrow{MN} = \overrightarrow{MA} + \overrightarrow{AN} = \mathbf{a} + \frac{4}{3}\mathbf{b}$

“Correct to 2 decimal places” tells you the equation does not factorise, so use the formula with $a = 1$, $b = 6$, $c = 1$:

$$\begin{aligned} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{36 - 4}}{2} \\ &= \frac{-6 \pm \sqrt{32}}{2} = \frac{-6 \pm 5.657}{2} \end{aligned}$$ $$x = -3 + 2.828 \quad\text{or}\quad x = -3 - 2.828$$

Answer: $x = -0.17$ or $x = -5.83$

Take out the common factor, then use the difference of two squares:

$$\begin{aligned} \frac{2x^2 - 50}{x+5} &= \frac{2(x^2 - 25)}{x+5} \\ &= \frac{2(x+5)(x-5)}{x+5} = 2(x-5) \end{aligned}$$

Answer: $2(x - 5)$ or $2x - 10$

Use the correct symbols: rounded boxes for Start/Stop, parallelograms for input and output, a rectangle for a calculation and a diamond for the decision.

(i) $q = 4^3 - 5(4)^2 - 4 + 5 = 64 - 80 - 4 + 5 = -15$

(iii)(a) The solutions are where the curve cuts the $x$-axis ($y = 0$): $x = -1$, $x = 1$, $x = 5$.

(iii)(b) Draw the straight line $y = 2x - 15$ on the same axes and read the $x$-values where it meets the curve: $x \approx -1.9$, $2.2$, $4.7$.

$$\begin{aligned} \frac{dy}{dx} &= 3x^2 - 2x - 15 \\ \text{At } x = 1:\; m &= 3 - 2 - 15 = -14 \end{aligned}$$ $$\begin{aligned} y - (-12) &= -14(x - 1) \\ y + 12 &= -14x + 14 \end{aligned}$$

Answer: $y = -14x + 2$

(i) Two sides and the angle between them, so use the cosine rule. Note that $\cos 91.2^\circ$ is negative:

$$\begin{aligned} AC^2 &= 125^2 + 57^2 - 2(125)(57)\cos 91.2^\circ \\ &= 15\,625 + 3\,249 - 14\,250(-0.020942) \\ &= 19\,172.4 \end{aligned}$$ $$AC = \sqrt{19\,172.4} = 138.46$$

Answer: $AC = 138$ cm (3 s.f.)

(ii)

$$\text{Area} = \tfrac{1}{2} \times 125 \times 57 \times \sin 91.2^\circ = 3\,561.7$$

Answer: $3\,560\ \text{cm}^2$ (3 s.f.)

(iii) The shortest distance from B to AC is the perpendicular height $h$ with AC as the base:

$$\begin{aligned} \tfrac{1}{2} \times 138.46 \times h &= 3\,561.7 \\ h &= \frac{2 \times 3\,561.7}{138.46} = 51.4 \end{aligned}$$

Answer: $51.4$ cm

(b) $\tan^{-1}(0.7) = 35.0^\circ$. Tan is positive in the first and third quadrants; in the given range we need the third: $\theta = 180^\circ + 35.0^\circ$.

Answer: $\theta = 215.0^\circ$

(c) Dividing by a fraction is multiplying by its reciprocal:

$$\begin{aligned} &\frac{52x^2y^2}{5a^2b^2} \times \frac{10a^3b^2}{13x^2y^3} \\ &= \frac{520\,x^2y^2a^3b^2}{65\,a^2b^2x^2y^3} = \frac{8a}{y} \end{aligned}$$

Answer: $\dfrac{8a}{y}$

(a) Use the mid-point of each class as $x$:

Mid-point $x$	5	15	25	35	45	55	Total
$f$	4	16	18	20	10	2	70
$fx$	20	240	450	700	450	110	1 970
$fx^2$	100	3 600	11 250	24 500	20 250	6 050	65 750

$$\bar{x} = \frac{\Sigma fx}{\Sigma f} = \frac{1970}{70} = 28.14$$ $$\begin{aligned} \text{SD} &= \sqrt{\frac{\Sigma fx^2}{\Sigma f} - \bar{x}^2} = \sqrt{\frac{65\,750}{70} - 28.14^2} \\ &= \sqrt{147.27} = 12.14 \end{aligned}$$

Answer: SD ≈ 12.1 years

(b)(i) Divide each cumulative frequency by 70:

Age (years)	≤ 0	≤ 10	≤ 20	≤ 30	≤ 40	≤ 50	≤ 60
Cumulative frequency	0	4	20	38	58	68	70
Relative cumulative frequency	0.00	0.06	0.29	0.54	0.83	0.97	1.00

(ii) and (iii) Plot each value at the upper class boundary (10, 20, 30…), not at the mid-point.

Read across from 0.25 and 0.75: $Q_1 \approx 18.6$ years and $Q_3 \approx 36.8$ years.

Answer: IQR $= Q_3 - Q_1 \approx 36.8 - 18.6 \approx 18$ years

(a) $x \ge 100,\quad y \ge 50,\quad x + y \le 600,\quad y \le 2x$

(b)

(c) Test the profit $P = 10x + 15y$ at each vertex of the wanted region:

Vertex	(100, 50)	(100, 200)	(550, 50)	(200, 400)
Profit	K1 750	K4 000	K6 250	K8 000

Answer: 200 type A bricks and 400 type B bricks

(d) $10(200) + 15(400) = 2\,000 + 6\,000$. Answer: K8 000.00

Read the coordinates first: A(−3, 1), B(−2, 4), C(−4, 3); A₁(1, 3), B₁(4, 2), C₁(3, 4).

(a) Each point follows $(x, y) \to (y, -x)$, for example A(−3, 1) → (1, 3).
Answer: a rotation of 90° clockwise (−90°) about the origin (0, 0). A full description needs all three: rotation, angle with direction, and centre.

(b) A₂(6, −2) = −2 × A(−3, 1), B₂(4, −8) = −2 × B(−2, 4), C₂(8, −6) = −2 × C(−4, 3).
Answer: centre (0, 0), scale factor −2

(c)(i) A₃(−1, 1), B₃(6, 4), C₃(2, 3): the $y$-coordinates are unchanged and $x' = x + ky$. Using B: $-2 + 4k = 6$, so $k = 2$ (check with A: $-3 + 2(1) = -1$ ✓).

Answer: $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$

(ii) Answer: shear factor $= 2$

(d)

$$\begin{aligned} &\begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} -3 & -2 & -4 \\ 1 & 4 & 3 \end{pmatrix} \\ &= \begin{pmatrix} -9 & -6 & -12 \\ 1 & 4 & 3 \end{pmatrix} \end{aligned}$$

Answer: A₄(−9, 1), B₄(−6, 4), C₄(−12, 3)

(i) Let the small cone that was removed have height $h$. The small and full cones are similar, so heights are in the same ratio as radii (3 : 12):

$$\begin{aligned} \frac{h}{h + 15} &= \frac{6}{24} \\ 24h &= 6h + 90 \\ 18h &= 90 \;\Rightarrow\; h = 5 \end{aligned}$$

Answer: height of the full cone $= 5 + 15 = 20$ cm

(ii) Volume of frustum = volume of full cone − volume of small cone:

$$\begin{aligned} V &= \tfrac{1}{3}\pi R^2 H - \tfrac{1}{3}\pi r^2 h \\ &= \tfrac{1}{3} \times 3.142 \times (12^2 \times 20 - 3^2 \times 5) \end{aligned}$$ $$\begin{aligned} &= \tfrac{1}{3} \times 3.142 \times (2880 - 45) \\ &= \tfrac{1}{3} \times 3.142 \times 2835 = 2969.19 \end{aligned}$$

Answer: $2\,970\ \text{cm}^3$ (3 s.f.)

(i)

(ii)(a) P and Q are on the same meridian (41°E), a great circle. One is north and one is south of the equator, so add the latitudes: $\theta = 5^\circ + 75^\circ = 80^\circ$.

$$\begin{aligned} PQ &= \frac{80}{360} \times 2 \times 3.142 \times 3437 \\ &= 4\,799.6 \text{ nm} \end{aligned}$$

Answer: PQ ≈ 4 800 nm. (Quick check: 1° along a great circle is 60 nm, and $80 \times 60 = 4\,800$.)

(ii)(b) Q and R are on the same parallel (75°S), a small circle of radius $R\cos 75^\circ$. One is east and one is west of Greenwich, so add the longitudes: $\theta = 41^\circ + 19^\circ = 60^\circ$.

$$\begin{aligned} QR &= \frac{60}{360} \times 2 \times 3.142 \times 3437 \times \cos 75^\circ \\ &= 3\,599.7 \times 0.2588 = 931.6 \text{ nm} \end{aligned}$$

Answer: QR ≈ 932 nm