GCE 2026 Mathematics Paper 1: Worked Solutions

A power of $\frac{1}{3}$ means the cube root. Write both numbers as cubes:

$$\left(\frac{27}{64}\right)^{\frac{1}{3}} = \left(\frac{3^3}{4^3}\right)^{\frac{1}{3}} = \frac{3}{4}$$

Answer: $\dfrac{3}{4}$

Group the terms in pairs and take out a common factor from each pair:

$$pq - 3p - q^2 + 3q = p(q - 3) - q(q - 3) = (q - 3)(p - q)$$

Answer: $(q - 3)(p - q)$

$$8x + 4 - 6x + 9 = 2x + 13$$

Answer: $2x + 13$

$$2x^2 = 72 \;\Rightarrow\; x^2 = 36 \;\Rightarrow\; x = \pm 6$$

Answer: $x = 6$ or $x = -6$

$(P \cup Q)'$ is everything outside both P and Q. Intersecting with R keeps only the part of R that is outside P and Q:

$$\text{Number of shares} = \frac{25\,250}{25} = 1\,010$$

Answer: 1 010 shares

(a) $a = -11$ and $T_n = a + (n - 1)d$, so

$$T_7 = -11 + 6d = 13 \;\Rightarrow\; 6d = 24 \;\Rightarrow\; d = 4$$

(b)

$$T_n = -11 + 4(n - 1) = 4n - 15$$

Answer: (a) $d = 4$ (b) $4n - 15$. Check: $T_7 = 4(7) - 15 = 13$ ✓

(a) The primes are 2, 3 and 5 (1 is not prime). $P(\text{prime}) = \dfrac{3}{6} = \dfrac{1}{2}$

(b)

$$\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} + \begin{pmatrix} 6 \\ 5 \end{pmatrix} = \begin{pmatrix} 7 \\ 6 \end{pmatrix}$$

Answer: (a) $\dfrac{1}{2}$ (b) B(7, 6)

(a) The transpose swaps rows and columns: $\mathbf{A}^T = \begin{pmatrix} 3 & -1 \\ 4 & 0 \end{pmatrix}$

(b) Multiply each row of A by the column B:

$$\mathbf{AB} = \begin{pmatrix} 3(-1) + 4(3) \\ -1(-1) + 0(3) \end{pmatrix} = \begin{pmatrix} 9 \\ 1 \end{pmatrix}$$

(a) 0 is included because $0 \le x$, and 10 is excluded because $x < 10$. Answer: $\{0, 2, 4, 6, 8\}$

(b) Subtract the starting point from the image:

$$\mathbf{T} = \begin{pmatrix} 5 - (-4) \\ -4 - 1 \end{pmatrix} = \begin{pmatrix} 9 \\ -5 \end{pmatrix}$$

(a) A and B are on the same meridian, one north and one south of the equator, so add the latitudes: $30^\circ + 30^\circ = 60^\circ$. Along a great circle, $1^\circ = 60$ nm:

$$\text{distance} = 60 \times 60 = 3\,600 \text{ nm}$$

(b) The difference in longitude is $30^\circ$, and the earth turns $1^\circ$ every 4 minutes:

$$30 \times 4 = 120 \text{ min} = 2 \text{ h}$$

The Prime Meridian is east of B, so its time is ahead: $09\,35 + 2\text{ h} = 11\,35$.

Answer: (a) 3 600 nm (b) 11 35 hours

(a)

$$\text{relative error} = \frac{\text{absolute error}}{n} \;\Rightarrow\; 0.2 = \frac{5}{n} \;\Rightarrow\; n = \frac{5}{0.2} = 25$$

(b) Percentage error $= 0.2 \times 100\% = 20\%$

Answer: (a) $n = 25$ (b) $20\%$

(a) Angles ACB and ACD lie on a straight line, so $\cos A\hat{C}B = -\cos A\hat{C}D = \dfrac{12}{13}$. In the right-angled triangle this means BC = 12 and AC = 13 (in proportion), and $AB = \sqrt{13^2 - 12^2} = 5$.

$$\sin C\hat{A}B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{12}{13}$$

(b) Write $\dfrac{12}{x^3}$ as $12x^{-3}$ first, then integrate each term:

$$\int (12x^{-3} - 3x^2)\,dx = \frac{12x^{-2}}{-2} - \frac{3x^3}{3} + c = -\frac{6}{x^2} - x^3 + c$$

Answer: (a) $\dfrac{12}{13}$ (b) $-\dfrac{6}{x^2} - x^3 + c$

(a) OB = OC (radii), so triangle OBC is isosceles and angle OCB = 50°. Then angle BOC $= 180^\circ - 100^\circ = 80^\circ$.

$$B\hat{D}C = \tfrac{1}{2} \times 80^\circ = 40^\circ \quad \text{(angle at the centre is twice the angle at the circumference)}$$

(b) Angle BCD $= 50^\circ + 25^\circ = 75^\circ$. ABCD is a cyclic quadrilateral, so opposite angles add up to 180°:

$$B\hat{A}D = 180^\circ - 75^\circ = 105^\circ$$

(c) Angle CBT = angle BDC = 40° (alternate segment theorem). D, C and T lie on a straight line, so angle BCT $= 180^\circ - 75^\circ = 105^\circ$. In triangle BCT:

$$B\hat{T}D = 180^\circ - 40^\circ - 105^\circ = 35^\circ$$

Answer: (a) 40° (b) 105° (c) 35°

(a) For similar solids, the volume ratio is the cube of the length ratio:

$$\left(\frac{y}{20}\right)^3 = \frac{8\,000}{1\,000} = 8 \;\Rightarrow\; \frac{y}{20} = \sqrt[3]{8} = 2 \;\Rightarrow\; y = 40$$

(b) Gradient of AB $= \dfrac{6 - 0}{14 - 5} = \dfrac{2}{3}$. Perpendicular gradients multiply to −1, so the new gradient is $-\dfrac{3}{2}$.

$$y - 9 = -\frac{3}{2}(x - 2) \;\Rightarrow\; y = -\frac{3}{2}x + 12$$

Answer: (a) $y = 40$ (b) $y = -\dfrac{3}{2}x + 12$

$$p = \frac{kq}{r^2}$$

(a) $6 = \dfrac{18k}{9} = 2k \;\Rightarrow\; k = 3$

(b) $9 = \dfrac{3q}{25} \;\Rightarrow\; 3q = 225 \;\Rightarrow\; q = 75$

(c) $3 = \dfrac{3(25)}{r^2} \;\Rightarrow\; 3r^2 = 75 \;\Rightarrow\; r^2 = 25 \;\Rightarrow\; r = \pm 5$

Answer: (a) $k = 3$ (b) $q = 75$ (c) $r = 5$ or $r = -5$

(a) The plus shape has 4 vertical planes of symmetry (front-to-back, left-to-right and the two diagonals) and 1 horizontal plane through the middle of the layer of cubes.

Answer: 5 planes of symmetry

(b)

Start
    Enter h, r
    v = π × r × r × h
    Output v
Stop

(a)

$$\text{Area} = \frac{60}{360} \times \frac{22}{7} \times 10.5 \times 10.5 = 57.75 \text{ cm}^2$$

(b)

$$\left(\frac{3 + (-9)}{2},\ \frac{6 + 4}{2}\right) = (-3,\ 5)$$

Answer: (a) 57.75 cm² (b) (−3, 5)

(a) Let $y = \dfrac{2}{x - 9}$ and make $x$ the subject:

$$y(x - 9) = 2 \;\Rightarrow\; xy = 2 + 9y \;\Rightarrow\; x = \frac{2 + 9y}{y}$$

So $f^{-1}(x) = \dfrac{2 + 9x}{x}$.

(b) $fg(x)$ means “do $g$ first, then $f$”: replace $x$ in $f$ with $x + 5$.

$$fg(x) = \frac{2}{(x + 5) - 9} = \frac{2}{x - 4}$$

(c) $\dfrac{2}{x - 4} = 1 \;\Rightarrow\; x - 4 = 2 \;\Rightarrow\; x = 6$

(a) A back bearing differs by 180°: $060^\circ + 180^\circ = 240^\circ$.

(b) OQ = QP, so triangle OQP is isosceles and angle QPO = angle QOP = 55°. Then angle OQP $= 180^\circ - 55^\circ - 55^\circ = 70^\circ$. From Q, P is 70° anticlockwise from the direction of O:

$$240^\circ - 70^\circ = 170^\circ$$

Answer: (a) 240° (b) 170°

• The vertical line $x = 2$, with R to the right: $x \ge 2$
• The horizontal line $y = -1$, with R above it: $y \ge -1$
• The horizontal line $y = 5$, with R below it: $y \le 5$
• The sloping line through (4, 5) and (9, 0): gradient $= \dfrac{0 - 5}{9 - 4} = -1$, and it cuts the $y$-axis at 9, so $y = -x + 9$, that is $x + y = 9$. R is below it: $x + y \le 9$

Answer: $x \ge 2,\quad y \ge -1,\quad y \le 5,\quad x + y \le 9$

(a) Write both sides as powers of 2:

$$(2^2)^{3x} = 2^3 \;\Rightarrow\; 2^{6x} = 2^3 \;\Rightarrow\; 6x = 3 \;\Rightarrow\; x = \tfrac{1}{2}$$

(b)(i) At A and B, $y = 0$: $-2x + 4 = 0$ or $x + 5 = 0$, so $x = 2$ or $x = -5$. Answer: A(−5, 0), B(2, 0)

(ii) Expand: $y = -2x^2 - 10x + 4x + 20 = -2x^2 - 6x + 20$. The turning point is at

$$x = -\frac{b}{2a} = -\frac{-6}{2(-2)} = -\frac{3}{2}$$

(halfway between −5 and 2). Substitute:

$$y = -2\left(\tfrac{9}{4}\right) - 6\left(-\tfrac{3}{2}\right) + 20 = -4.5 + 9 + 20 = 24.5$$

Answer: maximum value of $y$ is 24.5

(a) $a = \dfrac{v - u}{t} = \dfrac{10 - 0}{8} = 1.25 \text{ m/s}^2$

(b) Distance = area under the graph. The shape is a trapezium with parallel sides 60 s and 100 s and height 10 m/s:

$$\tfrac{1}{2}(60 + 100) \times 10 = 800 \text{ m}$$

(c) The deceleration lasts $100 - 68 = 32$ s, so $a = -\dfrac{10}{32} = -\dfrac{5}{16} \text{ m/s}^2$. At $t = 84$ s the particle has been slowing down for $84 - 68 = 16$ s:

$$v = u + at = 10 - \frac{5}{16}(16) = 5 \text{ m/s}$$

Answer: (a) 1.25 m/s² (b) 800 m (c) 5 m/s