2026 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Step 1: Apply the negative index rule to invert the fraction:
\[ \left(\frac{27}{64}\right)^{-\frac{2}{3}} = \left(\frac{64}{27}\right)^{\frac{2}{3}} \]
Step 2: Take the cube root of the numerator and the denominator:
\[ \sqrt[3]{\frac{64}{27}} = \frac{4}{3} \]
Step 3: Square the result:
\[ \left(\frac{4}{3}\right)^2 = \frac{16}{9} \text{ (or } 1\frac{7}{9}\text{)} \]

M1 for \( \left(\frac{64}{27}\right)^{\frac{2}{3}} \) or \( \left(\frac{3}{4}\right)^{-2} \) or \( \frac{1}{\left(\frac{3}{4}\right)^2} \)
A1 for \( \frac{16}{9} \) or \( 1\frac{7}{9} \)

Rationalise the first term:
\[\frac{18}{\sqrt{6}} = \frac{18\sqrt{6}}{6} = 3\sqrt{6}\]

Simplify the second term:
\[\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}\]

Combine the terms:
\[3\sqrt{6} + 2\sqrt{6} = 5\sqrt{6}\]

M1 for rationalising \(\frac{18}{\sqrt{6}}\) to \(3\sqrt{6}\) or simplifying \(\sqrt{24}\) to \(2\sqrt{6}\)
A1 for \(5\sqrt{6}\) or \(k = 5\)

(a) Apply the power \(\frac{3}{2}\) to each factor:
\(64^{\frac{3}{2}} = (\sqrt{64})^3 = 8^3 = 512\)
\((x^6)^{\frac{3}{2}} = x^{6 \times \frac{3}{2}} = x^9\)
\((y^{-4})^{\frac{3}{2}} = y^{-4 \times \frac{3}{2}} = y^{-6}\)
Combining these gives \(512 x^9 y^{-6}\) or \(\dfrac{512x^9}{y^6}\).

(b) Express the right-hand side as a power of 3:
\(27 = 3^3\) and \(\sqrt{3} = 3^{\frac{1}{2}}\)
\(27\sqrt{3} = 3^3 \times 3^{\frac{1}{2}} = 3^{\frac{7}{2}}\)
\(\dfrac{1}{27\sqrt{3}} = 3^{-\frac{7}{2}}\)

Equating indices:
\(2n - 1 = -\dfrac{7}{2}\)
\(2n = 1 - \dfrac{7}{2} = -\dfrac{5}{2}\)
\(n = -\dfrac{5}{4} = -1.25\).

(a)
B1: \(512\) seen or evaluated correctly.
B1: \(x^9 y^{-6}\) or \(\dfrac{x^9}{y^6}\) correct.

(b)
M1: Expressing the RHS as \(3^{-\frac{7}{2}}\) or \(3^{-3.5}\) or setting up \(2n - 1 = -3 - \frac{1}{2}\).
A1: \(n = -\dfrac{5}{4}\) or \(-1.25\) cao.

(a) Find the vector from the centre \((1, 4)\) to \(A(3, 2)\):
\(\begin{pmatrix} 3 - 1 \\ 2 - 4 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix}\).
Multiply by the scale factor \(-2\):
\(-2 \begin{pmatrix} 2 \\ -2 \end{pmatrix} = \begin{pmatrix} -4 \\ 4 \end{pmatrix}\).
Add this vector to the centre of enlargement:
\(\begin{pmatrix} 1 \\ 4 \end{pmatrix} + \begin{pmatrix} -4 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 8 \end{pmatrix}\).
Thus, \(A' = (-3, 8)\).

(b) Tracking a general point \((x, y)\):
- Rotation \(90^\circ\) clockwise about \((0, 0)\): \((x, y) \to (y, -x)\).
- Reflection in the line \(x = 0\) (the \(y\)-axis): \((y, -x) \to (-y, -x)\).
The mapping \((x, y) \to (-y, -x)\) is a reflection in the line \(y = -x\).

(a)
M1: Vector from centre found as \(\begin{pmatrix} 2 \\ -2 \end{pmatrix}\) and multiplied by \(-2\) to give \(\begin{pmatrix} -4 \\ 4 \end{pmatrix}\), or correct graphical method.
A1: \((-3, 8)\) cao.

(b)
B1: Reflection
B1: in the line \(y = -x\)
(Note: award 0 marks if more than one transformation is named).

(a) By the alternate segment theorem, the angle between the tangent \(TA\) and chord \(AB\) is equal to the angle subtended by the chord in the alternate segment:
Angle \(BCA = 56^\circ\).
Reason: Alternate segment theorem.

(b) The angle subtended by an arc at the centre is twice the angle subtended at the circumference:
Angle \(AOB = 2 \times \text{Angle } BCA = 2 \times 56^\circ = 112^\circ\).

(c) Angles in the same segment subtended by chord \(AB\) are equal, so Angle \(ADB = \text{Angle } BCA = 56^\circ\).
Therefore:
Angle \(ADC = \text{Angle } ADB + \text{Angle } BDC = 56^\circ + 38^\circ = 94^\circ\).

(a)
B1: \(56^\circ\)
B1: alternate segment (theorem)

(b)
B1: \(112^\circ\)

(c)
B1: \(94^\circ\)

(a) Linear scale factor \(k = \dfrac{18}{12} = \dfrac{3}{2}\).
Area scale factor \(= k^2 = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}\).
Surface area of \(B = 140 \times \dfrac{9}{4} = 35 \times 9 = 315\text{ cm}^2\).

(b) Volume (and mass) scale factor \(= k^3 = \left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8}\).
Mass of \(B = 2.7\text{ kg} = 2700\text{ g}\).
Mass of \(A = \dfrac{2700}{\frac{27}{8}} = 2700 \times \dfrac{8}{27} = 100 \times 8 = 800\text{ g}\).

(a)
M1: \(140 \times \left(\dfrac{18}{12}\right)^2\) oe
A1: \(315\)

(b)
M1: \(2700 \div \left(\dfrac{18}{12}\right)^3\) or \(2.7 \times \left(\dfrac{12}{18}\right)^3 \times 1000\) oe
A1: \(800\) cao

(a) Express all terms with base 2:
\(8^{x+1} = (2^3)^{x+1} = 2^{3(x+1)} = 2^{3x+3}\)
\(4 \times 2^{5-x} = 2^2 \times 2^{5-x} = 2^{2 + 5 - x} = 2^{7-x}\)

Equating the powers:
\(3x + 3 = 7 - x\)
\(4x = 4\)
\(x = 1\).

(b) Deal with the negative exponent by taking the reciprocal:
\(\left(\dfrac{16}{81}\right)^{-\frac{3}{4}} = \left(\dfrac{81}{16}\right)^{\frac{3}{4}}\)
Take the 4th root, then cube:
\(\sqrt[4]{\dfrac{81}{16}} = \dfrac{3}{2}\)
\(\left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8}\).

(a)
M1: Expressing both sides as powers of 2, e.g. \(2^{3x+3} = 2^{7-x}\) or \(3(x+1) = 2 + 5 - x\).
A1: \(x = 1\) cao.

(b)
M1: Inverting fraction to \(\left(\frac{81}{16}\right)^{\frac{3}{4}}\) or evaluating 4th root to get \(\left(\frac{2}{3}\right)^{-3}\) or \(\frac{3}{2}\).
A1: \(\dfrac{27}{8}\) or \(3\dfrac{3}{8}\) cao.

(a) Add the translation vector \(\begin{pmatrix} -3 \\ 1 \end{pmatrix}\) to each vertex of \(P\):
\((1, 2) \to (1 - 3, 2 + 1) = (-2, 3)\)
\((4, 2) \to (4 - 3, 2 + 1) = (1, 3)\)
\((4, 5) \to (4 - 3, 5 + 1) = (1, 6)\)

(b) Comparing the coordinates of corresponding vertices:
\((1, 2) \to (-2, 1)\)
\((4, 2) \to (-2, 4)\)
\((4, 5) \to (-5, 4)\)
The rule is \((x, y) \to (-y, x)\), which corresponds to a rotation of \(90^\circ\) anticlockwise (or \(270^\circ\) clockwise) about the origin \((0, 0)\).

(a)
B1: Correct coordinates for all 3 vertices: \((-2, 3)\), \((1, 3)\), \((1, 6)\).

(b)
B1: Rotation of \(90^\circ\) anticlockwise (or \(270^\circ\) clockwise).
B1: Centre \((0, 0)\) or origin.
(Note: Award 0 for part (b) if more than one transformation is mentioned).

(a) Let the interior angle be \(I\) and the exterior angle be \(E\).
We know that \(I + E = 180^\circ\) and \(I - E = 140^\circ\).
Subtracting the two equations:
\(2E = 180^\circ - 140^\circ = 40^\circ \implies E = 20^\circ\).
The sum of the exterior angles of any polygon is \(360^\circ\).
Number of sides \(= \dfrac{360^\circ}{E} = \dfrac{360^\circ}{20^\circ} = 18\).

(b) Exterior angle \(= 180^\circ - 156^\circ = 24^\circ\).
\(n = \dfrac{360^\circ}{24^\circ} = 15\).

(a)
M1: \(E + (E + 140) = 180\) or \(E = 20^\circ\) or \(I = 160^\circ\).
A1: 18 cao.

(b)
M1: Exterior angle \(= 180 - 156 = 24^\circ\) or \(\dfrac{(n-2) \times 180}{n} = 156\).
A1: 15 cao.

Step 1: Rationalise the denominator of \(\frac{14}{3 - \sqrt{2}}\):
\[\frac{14(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})} = \frac{14(3 + \sqrt{2})}{3^2 - (\sqrt{2})^2} = \frac{14(3 + \sqrt{2})}{9 - 2} = \frac{14(3 + \sqrt{2})}{7}\]
\[= 2(3 + \sqrt{2}) = 6 + 2\sqrt{2}\]

Step 2: Simplify \(\sqrt{32}\):
\[\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}\]

Step 3: Combine the terms:
\[(6 + 2\sqrt{2}) - 4\sqrt{2} = 6 - 2\sqrt{2}\]

M1 for multiplying numerator and denominator by \(3 + \sqrt{2}\)
A1 for obtaining \(6 + 2\sqrt{2}\) or \(\frac{42 + 14\sqrt{2}}{7}\)
B1 for \(\sqrt{32} = 4\sqrt{2}\)
A1 for final answer \(6 - 2\sqrt{2}\) cao (must have \(a = 6\), \(b = -2\))

Step 1: Identify that the common denominator is \((2x - 1)(2x + 1) = 4x^2 - 1\).

Step 2: Express the fractions over the common denominator:
\[\frac{3(2x + 1) - 2(2x - 1) - (2x + 5)}{(2x - 1)(2x + 1)}\]

Step 3: Expand the terms in the numerator:
\[3(2x + 1) = 6x + 3\]
\[-2(2x - 1) = -4x + 2\]
\[-(2x + 5) = -2x - 5\]

Step 4: Collect and simplify the numerator terms:
\[(6x - 4x - 2x) + (3 + 2 - 5) = 0x + 0 = 0\]

Step 5: Conclude the proof:
\[\frac{0}{(2x - 1)(2x + 1)} = 0\]

M1 for recognising \(4x^2 - 1 = (2x - 1)(2x + 1)\) and writing expressions over a common denominator \((2x - 1)(2x + 1)\)
M1 for correct expansion of brackets in numerator: \(6x + 3 - 4x + 2 - (2x + 5)\) (allow 1 sign error)
A1 for correct simplified numerator showing \(6x - 4x - 2x + 3 + 2 - 5 = 0\)
A1 for fully correct and convincing conclusion with no steps omitted

Step 1: Find the midpoint \(M\) of \(PQ\):
\[M = \left(\frac{-3 + 5}{2}, \frac{8 + (-4)}{2}\right) = \left(\frac{2}{2}, \frac{4}{2}\right) = (1, 2)\]

Step 2: Find the gradient of line segment \(PQ\):
\[m_{PQ} = \frac{-4 - 8}{5 - (-3)} = \frac{-12}{8} = -\frac{3}{2}\]

Step 3: Find the gradient of the perpendicular bisector:
\[m_{\perp} = -\frac{1}{m_{PQ}} = \frac{2}{3}\]

Step 4: Use the point-slope formula with midpoint \((1, 2)\) and gradient \(\frac{2}{3}\):
\[y - 2 = \frac{2}{3}(x - 1)\]
\[3(y - 2) = 2(x - 1)\]
\[3y - 6 = 2x - 2\]
\[2x - 3y + 4 = 0\]

B1 for midpoint \((1, 2)\)
M1 for gradient of \(PQ = \frac{-4 - 8}{5 - (-3)} = -\frac{3}{2}\)
M1 for perpendicular gradient \(m = \frac{2}{3}\) (using negative reciprocal of their gradient)
M1 for substituting their midpoint and their perpendicular gradient into a linear equation \(y - y_1 = m(x - x_1)\) or \(y = mx + c\)
A1 for \(2x - 3y + 4 = 0\) (or any non-zero integer multiple with \(a > 0\)) cao

(a) Let \(y = \frac{3x - 1}{x + 2}\):
\[y(x + 2) = 3x - 1\]
\[yx + 2y = 3x - 1\]
\[2y + 1 = 3x - yx\]
\[2y + 1 = x(3 - y)\]
\[x = \frac{2y + 1}{3 - y}\]
Therefore,
\[\mathrm{f}^{-1}(x) = \frac{2x + 1}{3 - x} \quad \left(\text{or } \frac{-2x - 1}{x - 3}\right)\]

(b) Set up \(\mathrm{g}(\mathrm{f}(x)) = 9\):
\[5 - 2\left(\frac{3x - 1}{x + 2}\right) = 9\]
\[-2\left(\frac{3x - 1}{x + 2}\right) = 4\]
\[\frac{3x - 1}{x + 2} = -2\]
\[3x - 1 = -2(x + 2)\]
\[3x - 1 = -2x - 4\]
\[5x = -3\]
\[x = -\frac{3}{5}\]

(a) M1 for multiplying by \(x + 2\) to get \(y(x + 2) = 3x - 1\) (or \(x(y + 2) = 3y - 1\))
M1 for isolating terms in \(x\) (or \(y\)): \(x(3 - y) = 2y + 1\)
A1 for \(\mathrm{f}^{-1}(x) = \frac{2x + 1}{3 - x}\) or \(\frac{-2x - 1}{x - 3}\)
(b) M1 for setting up \(5 - 2\left(\frac{3x-1}{x+2}\right) = 9\) or using \(\mathrm{f}(x) = \mathrm{g}^{-1}(9)\)
A1 for \(x = -\frac{3}{5}\) or \(-0.6\) cao

Step 1: Use the formula for the area of a triangle, \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\):
\[\frac{1}{2} (\sqrt{5} - 1) h = 3\sqrt{5} + 5\]

Step 2: Multiply by 2 to isolate \((\sqrt{5} - 1) h\):
\[(\sqrt{5} - 1) h = 2(3\sqrt{5} + 5) = 6\sqrt{5} + 10\]

Step 3: Express \(h\) and rationalise the denominator:
\[h = \frac{10 + 6\sqrt{5}}{\sqrt{5} - 1}\]
Multiply numerator and denominator by \(\sqrt{5} + 1\):
\[h = \frac{(10 + 6\sqrt{5})(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)}\]

Step 4: Expand numerator and denominator:
\[\text{Numerator} = 10\sqrt{5} + 10 + 6(\sqrt{5})^2 + 6\sqrt{5} = 10\sqrt{5} + 10 + 30 + 6\sqrt{5} = 40 + 16\sqrt{5}\]
\[\text{Denominator} = (\sqrt{5})^2 - 1^2 = 5 - 1 = 4\]

Step 5: Divide to simplify:
\[h = \frac{40 + 16\sqrt{5}}{4} = 10 + 4\sqrt{5}\text{ cm}\]

M1 for setting up equation \(\frac{1}{2}(\sqrt{5} - 1)h = 3\sqrt{5} + 5\) or \(h = \frac{2(3\sqrt{5} + 5)}{\sqrt{5} - 1}\)
M1 for multiplying numerator and denominator by \(\sqrt{5} + 1\)
M1 for correct expansion of numerator to \(40 + 16\sqrt{5}\) or denominator to 4
A1 for \(10 + 4\sqrt{5}\) (accept \(p = 10, q = 4\)) cao

The upper bound for the length is \(8.4 + 0.05 = 8.45\text{ cm}\).
The upper bound for the width is \(5.2 + 0.05 = 5.25\text{ cm}\).

Upper bound of perimeter:
\[ 2 \times (8.45 + 5.25) = 2 \times 13.7 = 27.4\text{ cm} \]

M1 for upper bounds \(8.45\) and \(5.25\) seen
A1 for 27.4 cao

The area of a sector is given by:
\[ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 \]
Substitute \(\theta = 112^\circ\) and \(r = 7.5\text{ cm}\):
\[ \text{Area} = \frac{112}{360} \times \pi \times 7.5^2 = \frac{112}{360} \times \pi \times 56.25 \approx 54.97787... \]
Rounding to 3 significant figures gives \(55.0\text{ cm}^2\).

M1 for \(\frac{112}{360} \times \pi \times 7.5^2\) oe
A1 for 55.0 or 54.98 or answer in range [54.97, 55.0]

Find the highest common factor of \(6x^2\) and \(15xy\), which is \(3x\).

Factorise:
\[ 6x^2 - 15xy = 3x(2x - 5y) \]

B2 for \(3x(2x - 5y)\)
or B1 for partial factorisation such as \(3(2x^2 - 5xy)\) or \(x(6x - 15y)\)

The magnitude of a column vector \(\begin{pmatrix} x \\ y \end{pmatrix}\) is \(\sqrt{x^2 + y^2}\).
\[ |\mathbf{v}| = \sqrt{(-7)^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \]

M1 for \((-7)^2 + 24^2\) or \(\sqrt{(-7)^2 + 24^2}\) oe
A1 for 25 cao

The gradient of \(L_1\) is \(4\).
The gradient of \(L_2\) is \(m = -\frac{1}{4} = -0.25\).

Using point-slope form with \((8, 3)\):
\[ y - 3 = -0.25(x - 8) \]
\[ y - 3 = -0.25x + 2 \implies y = -0.25x + 5 \]

M1 for gradient of perpendicular line \(-\frac{1}{4}\) or \(-0.25\) used
A1 for \(y = -0.25x + 5\) or \(y = -\frac{1}{4}x + 5\) oe

Differentiate \(y\) with respect to \(x\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 2(3x^2) - 9(2x) + 0 = 6x^2 - 18x \]
Substitute \(x = 4\):
\[ \text{Gradient} = 6(4)^2 - 18(4) = 6(16) - 72 = 96 - 72 = 24 \]

M1 for derivative \(6x^2 - 18x\) (at least one term correct)
A1 for 24 cao

Use the compound growth formula \(V = P(1 + r)^n\):
\[ V = 3200 \times (1 + 0.045)^6 = 3200 \times (1.045)^6 \]
\[ V = 3200 \times 1.302260... \approx 4167.23... \]
Rounding to the nearest dollar gives \(\$4167\).

M1 for \(3200 \times 1.045^6\) oe
A1 for 4167 (accept 4167.23 or 4167.2 if not rounded, or rounded to nearest dollar)

In triangle \(ABC\), relative to angle \(BAC\):
- Opposite side = \(BC = 9.5\text{ cm}\)
- Adjacent side = \(AB = 6.8\text{ cm}\)

\[ \tan(BAC) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{9.5}{6.8} \]
\[ \text{Angle } BAC = \tan^{-1}\left(\frac{9.5}{6.8}\right) = 54.407...^\circ \]
Rounding to 1 decimal place gives \(54.4^\circ\).

M1 for \(\tan(BAC) = \frac{9.5}{6.8}\) or \(\tan^{-1}\left(\frac{9.5}{6.8}\right)\) oe
A1 for 54.4 or 54.41...

(a)
1. Total rod length: \(2x + 5y \le 90\)
2. Minimum type \(P\) brackets: \(x \ge 12\)
3. More than half condition: \(y > 0.5x\) (or \(2y > x\))

(b)
We need to find integers \(x, y\) such that \(x \ge 12\), \(y > 0.5x\), and \(2x + 5y \le 90\) to maximise \(x + y\).

Rearranging the first inequality gives \(5y \le 90 - 2x \implies y \le \frac{90 - 2x}{5}\).
Testing possible integer values of \(x\ge 12\):
- If \(x = 17\): \(y \le \frac{90 - 34}{5} = 11.2\). Since \(y > 8.5\), possible integers for \(y\) are \(9, 10, 11\). Max \(x + y = 17 + 11 = 28\).
- If \(x = 18\): \(y \le \frac{90 - 36}{5} = 10.8\). Since \(y > 9\), possible integer is \(y = 10\). Max \(x + y = 18 + 10 = 28\).
- If \(x = 19\): \(y \le \frac{90 - 38}{5} = 10.4\). Since \(y > 9.5\), possible integer is \(y = 10\). Total \(x + y = 19 + 10 = 29\).
- If \(x = 20\): \(y \le \frac{90 - 40}{5} = 10\). But \(y > 10\), so no integer \(y\) exists.

Thus, the maximum total number of brackets is \(29\) (achieved at \(x = 19, y = 10\)).

(a)
B1 for \(2x + 5y \le 90\) oe
B1 for \(x \ge 12\) and \(y > 0.5x\) (or \(2y > x\))

(b)
M1 for systematic testing of integer pairs \((x, y)\) satisfying all three inequalities
A1 for 29 cao

(a)
Deceleration is \(0.5\text{ m/s}^2\).
\(\text{Time taken} = \frac{\text{change in speed}}{\text{deceleration}} = \frac{V - 0}{0.5} = 2V\text{ seconds}\).

(b)
Total distance is \(3.25\text{ km} = 3250\text{ m}\).
Distance in stage 1 (acceleration): \(\frac{1}{2} \times 30 \times V = 15V\)
Distance in stage 2 (constant speed): \(90 \times V = 90V\)
Distance in stage 3 (deceleration): \(\frac{1}{2} \times 2V \times V = V^2\)

Total distance:
\(15V + 90V + V^2 = 3250\)
\(V^2 + 105V - 3250 = 0\)
\((V - 25)(V + 130) = 0\)

Since speed \(V > 0\), \(V = 25\).

(a)
B1 for \(\frac{V}{0.5} = 2V\) with clear explanation

(b)
M1 for setting up distance expression: \(15V + 90V + \frac{1}{2}(2V)(V)\) oe
M1 for quadratic equation \(V^2 + 105V - 3250 = 0\) correctly simplified and factorised (or correct quadratic formula applied)
A1 for 25 cao

(a)
Let \(y = \frac{2x - 1}{x + 3}\).
\(y(x + 3) = 2x - 1\)
\(yx + 3y = 2x - 1\)
\(3y + 1 = 2x - yx\)
\(3y + 1 = x(2 - y)\)
\(x = \frac{3y + 1}{2 - y}\)

Therefore, \(\mathrm{f}^{-1}(x) = \frac{3x + 1}{2 - x}\).

(b)
Method 1: Using \(\mathrm{f}^{-1}\):
\(\mathrm{f}\mathrm{g}(x) = 1 \implies \mathrm{g}(x) = \mathrm{f}^{-1}(1)\)
\(\mathrm{f}^{-1}(1) = \frac{3(1) + 1}{2 - 1} = \frac{4}{1} = 4\)
\(4x - 5 = 4 \implies 4x = 9 \implies x = 2.25\)

Method 2: Directly substituting \(\mathrm{g}(x)\):
\(\mathrm{f}(4x - 5) = \frac{2(4x - 5) - 1}{(4x - 5) + 3} = \frac{8x - 11}{4x - 2} = 1\)
\(8x - 11 = 4x - 2 \implies 4x = 9 \implies x = 2.25\).

(a)
M1 for multiplying by denominator and isolating terms in \(x\): \(yx - 2x = -3y - 1\) or \(2x - yx = 3y + 1\)
A1 for \(\frac{3x + 1}{2 - x}\) oe

(b)
M1 for setting \(\frac{8x - 11}{4x - 2} = 1\) or \(4x - 5 = \mathrm{f}^{-1}(1)\)
A1 for \(2.25\) or \(\frac{9}{4}\)

(a)
Total frequency is \(160\).
\(32 + 48 + 35 + k + 15 = 160\)
\(130 + k = 160 \implies k = 30\).

(b)
Frequency density \(\text{FD} = \frac{\text{Frequency}}{\text{Class width}}\).
For interval \(140 < m \le 160\):
\(\text{Class width} = 160 - 140 = 20\)
\(\text{FD} = \frac{48}{20} = 2.4\)

The height of the bar is \(7.2\text{ cm}\), so the scale factor is:
\(\text{Scale} = \frac{7.2\text{ cm}}{2.4} = 3\text{ cm per unit of FD}\).

For interval \(170 < m \le 200\):
\(\text{Class width} = 200 - 170 = 30\)
\(\text{Frequency} = k = 30\)
\(\text{FD} = \frac{30}{30} = 1.0\)

\(\text{Height of bar} = 1.0 \times 3\text{ cm} = 3\text{ cm}\).

(a)
B1 for 30 cao

(b)
M1 for finding frequency density of \(140 < m \le 160\) as \(\frac{48}{20} = 2.4\)
M1 for finding frequency density of \(170 < m \le 200\) as \(\frac{30}{30} = 1.0\) and using proportionality with \(7.2\)
A1 for 3 cao

(a)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 2(3x^2) - 9(2x) - 24 = 6x^2 - 18x - 24\).

(b)
At stationary points, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(6x^2 - 18x - 24 = 0\)
\(6(x^2 - 3x - 4) = 0\)
\((x - 4)(x + 1) = 0\)
\(x = -1\) or \(x = 4\).

When \(x = -1\):
\(y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 15 = -2 - 9 + 24 + 15 = 28\)
Coordinates: \((-1, 28)\).

When \(x = 4\):
\(y = 2(4)^3 - 9(4)^2 - 24(4) + 15 = 128 - 144 - 96 + 15 = -97\)
Coordinates: \((4, -97)\).

Using the second derivative to determine nature:
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x - 18\)
At \(x = -1\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(-1) - 18 = -30 < 0\), so \((-1, 28)\) is a local maximum.
At \(x = 4\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(4) - 18 = 30 > 0\), so \((4, -97)\) is a local minimum.

(a)
B1 for \(6x^2 - 18x - 24\)

(b)
M1 for equating their derivative to 0 and solving to find \(x = -1\) and \(x = 4\)
M1 for substituting both \(x\) values into the original curve equation to find \(y = 28\) and \(y = -97\)
A1 for identifying \((-1, 28)\) as maximum and \((4, -97)\) as minimum with correct justification (second derivative or gradient test)

(a)
Time for outward journey = \(\frac{36}{x}\text{ hours}\).
Time for return journey = \(\frac{36}{x - 3}\text{ hours}\).
\(24\text{ minutes} = \frac{24}{60}\text{ hours} = \frac{2}{5}\text{ hours}\).

\(\frac{36}{x - 3} - \frac{36}{x} = \frac{2}{5}\)
\(\frac{36x - 36(x - 3)}{x(x - 3)} = \frac{2}{5}\)
\(\frac{108}{x^2 - 3x} = \frac{2}{5}\)
\(2(x^2 - 3x) = 540\)
\(2x^2 - 6x - 540 = 0\)
Dividing by 2:
\(x^2 - 3x - 270 = 0\).

(b)
\(x^2 - 3x - 270 = 0\)
\((x - 18)(x + 15) = 0\)
Since speed must be positive, \(x = 18\).

Return speed = \(18 - 3 = 15\text{ km/h}\).
Time for return journey = \(\frac{36}{15} = 2.4\text{ hours}\).
\(0.4\text{ hours} = 0.4 \times 60 = 24\text{ minutes}\).
Total time = \(2\text{ hours } 24\text{ minutes}\).

(a)
M1 for setting up \(\frac{36}{x - 3} - \frac{36}{x} = \frac{24}{60}\) oe
M1 for clearing fractions correctly: \(36x - 36(x - 3) = \frac{2}{5}(x^2 - 3x)\) or equivalent
A1 for complete rigorous algebraic progression leading to \(x^2 - 3x - 270 = 0\) without omitted steps

(b)
B1 for \(2\text{ hours } 24\text{ minutes}\) (or \(2.4\text{ hours}\) correctly converted to \(2\text{ h } 24\text{ min}\))

(a)
For integer values of \(y\ge 2\), the conditions require \(0 \le x \le y + 1\) and \(x \le \frac{24 - 3y}{2}\):
- For \(y = 2\): \(x \le 3\) and \(x \le 9 \implies x \in \{0, 1, 2, 3\}\) (4 points)
- For \(y = 3\): \(x \le 4\) and \(x \le 7.5 \implies x \in \{0, 1, 2, 3, 4\}\) (5 points)
- For \(y = 4\): \(x \le 5\) and \(x \le 6 \implies x \in \{0, 1, 2, 3, 4, 5\}\) (6 points)
- For \(y = 5\): \(x \le 6\) and \(x \le 4.5 \implies x \in \{0, 1, 2, 3, 4\}\) (5 points)
- For \(y = 6\): \(x \le 7\) and \(x \le 3 \implies x \in \{0, 1, 2, 3\}\) (4 points)
- For \(y = 7\): \(x \le 8\) and \(x \le 1.5 \implies x \in \{0, 1\}\) (2 points)
- For \(y = 8\): \(x \le 9\) and \(x \le 0 \implies x = 0\) (1 point)
- For \(y \ge 9\): \(x < 0\), no non-negative integer \(x\).

Total number of integer points = \(4 + 5 + 6 + 5 + 4 + 2 + 1 = 27\).

(b)
Evaluating \(3x + 4y\) at boundary integer points:
- \((0, 8) \implies 3(0) + 4(8) = 32\)
- \((1, 7) \implies 3(1) + 4(7) = 31\)
- \((3, 6) \implies 3(3) + 4(6) = 9 + 24 = 33\)
- \((4, 5) \implies 3(4) + 4(5) = 12 + 20 = 32\)
- \((5, 4) \implies 3(5) + 4(4) = 15 + 16 = 31\)

Maximum value is \(33\) at point \((3, 6)\).

(a)
M1 for listing or systematically counting valid integer \(x\)-values for each integer \(y\)
A1 for 27 cao

(b)
M1 for testing integer boundary coordinates near the top/right edges into \(3x + 4y\)
A1 for point \((3, 6)\) and maximum value \(33\)

(a)
\(\text{Frequency} = \text{Frequency density} \times \text{Class width}\).
- For \(0 < t \le 5\): \(4.8 \times 5 = 24\)
- For \(5 < t \le 10\): \(8.4 \times 5 = 42\)
- For \(10 < t \le 15\): \(9.2 \times 5 = 46\)
- For \(25 < t \le 40\): \(1.6 \times 15 = 24\)

Sum of known frequencies = \(24 + 42 + 46 + 24 = 136\).
Total number of customers is \(200\).
Frequency for \(15 < t \le 25 = 200 - 136 = 64\).

(b)
For \(15 < t \le 25\):
Class width = \(25 - 15 = 10\).
Frequency = \(64\).
Customers waiting between \(18\) and \(25\text{ minutes}\):
Width = \(25 - 18 = 7\).
\(\text{Estimated frequency} = \frac{7}{10} \times 64 = 44.8\).

Customers waiting between \(25\) and \(40\text{ minutes} = 24\).

Total customers waiting more than \(18\text{ minutes}\) = \(44.8 + 24 = 68.8\) (or \(69\)).

(a)
M1 for calculating the frequencies of the 4 given intervals: \(24, 42, 46, 24\)
A1 for complete calculation showing \(200 - (24 + 42 + 46 + 24) = 64\)

(b)
M1 for \(\frac{25 - 18}{10} \times 64\) (or \(44.8\))
A1 for \(68.8\) or \(69\)

(a) Let \(y = 3x - 5\).
Rearrange to make \(x\) the subject:
\[y + 5 = 3x\]
\[x = \frac{y + 5}{3}\]
Therefore, \(\mathrm{f}^{-1}(x) = \frac{x + 5}{3}\).

(b) First find the composite function \(\mathrm{gf}(x)\):
\[\mathrm{gf}(x) = \mathrm{g}(3x - 5) = \frac{8}{2(3x - 5) + 1} = \frac{8}{6x - 10 + 1} = \frac{8}{6x - 9}\]
Set \(\mathrm{gf}(x) = 2\):
\[\frac{8}{6x - 9} = 2\]
\[8 = 2(6x - 9)\]
\[8 = 12x - 18\]
\[12x = 26\]
\[x = \frac{26}{12} = \frac{13}{6} \approx 2.17\]

Alternatively, solve \(\mathrm{g}(u) = 2\):
\[\frac{8}{2u + 1} = 2 \implies 2u + 1 = 4 \implies 2u = 3 \implies u = 1.5\]
Then solve \(\mathrm{f}(x) = 1.5\):
\[3x - 5 = 1.5 \implies 3x = 6.5 \implies x = \frac{6.5}{3} = \frac{13}{6}\]

(a)
M1 for correct first step in finding inverse: \(x = 3y - 5\) or \(y + 5 = 3x\) or \(\frac{y+5}{3}\)
A1 for \(\frac{x + 5}{3}\) or \(\frac{1}{3}(x + 5)\) oe

(b)
M1 for setting up \(\frac{8}{2(3x - 5) + 1} = 2\) or finding \(\mathrm{g}^{-1}(2) = 1.5\) and setting \(3x - 5 = 1.5\)
A1 for \(\frac{13}{6}\) or \(2\frac{1}{6}\) or \(2.17\) or \(2.166\dots\)

(a) In the horizontal base \(ABCD\), calculate the diagonal \(AC\):
\[AC^2 = AB^2 + BC^2 = 16^2 + 12^2 = 256 + 144 = 400\]
\[AC = \sqrt{400} = 20\text{ cm}\]
Since \(M\) is the centre of the base, \(AM = \frac{1}{2}AC = 10\text{ cm}\).
In right-angled triangle \(VMA\):
\[VM = \sqrt{VA^2 - AM^2} = \sqrt{18^2 - 10^2} = \sqrt{324 - 100} = \sqrt{224} \approx 14.9666\text{ cm}\]
Rounded to 3 significant figures, \(VM = 15.0\text{ cm}\) (or \(14.97\text{ cm}\) to 4 s.f., exact \(4\sqrt{14}\)).

(b) Let \(N\) be the midpoint of \(BC\). Then \(VN \perp BC\) and \(MN \perp BC\).
The distance \(MN = \frac{1}{2}AB = 8\text{ cm}\).
In right-angled triangle \(VMN\), the angle between face \(VBC\) and the base is \(\angle VNM\):
\[\tan(\angle VNM) = \frac{VM}{MN} = \frac{\sqrt{224}}{8} = \frac{14.9666}{8} \approx 1.8708\]
\[\angle VNM = \tan^{-1}(1.8708) \approx 61.874^\circ\]
Correct to 1 decimal place, the angle is \(61.9^\circ\).

(a)
M1 for \(AM = \frac{1}{2}\sqrt{16^2 + 12^2} = 10\) or \(AC = 20\)
A1 for 15.0 or 14.96 to 14.97 or \(\sqrt{224}\) or \(4\sqrt{14}\)

(b)
B1 for identifying/calculating \(MN = 8\) or \(VN = \sqrt{18^2 - 6^2} = \sqrt{288}\)
M1 for \(\tan(\theta) = \frac{\text{their } VM}{8}\) or \(\cos(\theta) = \frac{8}{\sqrt{288}}\) or \(\sin(\theta) = \frac{\text{their } VM}{\sqrt{288}}\)
A1 for 61.9 or 61.87 to 61.88

(a) Number of years from 1 January 2015 to 1 January 2023 is \(t = 2023 - 2015 = 8\).
\[\text{Value} = 14000 \times (1 + 0.045)^8 = 14000 \times 1.045^8 \approx 19909.41\]
To the nearest dollar, the value is \(\$19\,909\).

(b) Let \(t\) be the number of years after 2015.
Painting value: \(P(t) = 14000 \times 1.045^t\)
Sculpture value: \(S(t) = 22000 \times (1 - 0.03)^t = 22000 \times 0.97^t\)
We require \(P(t) > S(t)\):
\[14000 \times 1.045^t > 22000 \times 0.97^t\]
\[\left(\frac{1.045}{0.97}\right)^t > \frac{22000}{14000} = \frac{11}{7}\]
\[(1.07732)^t > 1.57143\]
Evaluating for integer values of \(t\):
- For \(t = 6\): \(P(6) = 14000 \times 1.045^6 \approx 18231.67\) and \(S(6) = 22000 \times 0.97^6 \approx 18325.43\) (Painting < Sculpture)
- For \(t = 7\): \(P(7) = 14000 \times 1.045^7 \approx 19052.09\) and \(S(7) = 22000 \times 0.97^7 \approx 17775.67\) (Painting > Sculpture)
Therefore, the condition is first met after \(t = 7\) years.
Year = \(2015 + 7 = 2022\).

(a)
M1 for \(14000 \times 1.045^8\) oe
A1 for 19 909 cao

(b)
M1 for setting up inequality or equation \(14000 \times 1.045^t = 22000 \times 0.97^t\) oe
M1 for systematic trial of values around \(t = 6\) and \(t = 7\) or solving \(t > 6.07\)
A1 for 2022 cao

(a) The radius \(r = 3.4\text{ cm}\) correct to 1 decimal place has bounds:
\[3.35 \le r < 3.45\]
Lower bound of \(r = 3.35\text{ cm}\).

(b) Find the bounds for each measurement:
- Mass \(m = 780\text{ g}\) (correct to 2 s.f., nearest 10): Upper bound \(m_{\text{UB}} = 785\text{ g}\)
- Height \(h = 8.5\text{ cm}\) (correct to 1 d.p.): Lower bound \(h_{\text{LB}} = 8.45\text{ cm}\)
- Radius \(r = 3.4\text{ cm}\): Lower bound \(r_{\text{LB}} = 3.35\text{ cm}\)

To maximise \(D = \frac{m}{\pi r^2 h}\), use the maximum numerator and minimum denominator:
\[D_{\text{UB}} = \frac{m_{\text{UB}}}{\pi \times (r_{\text{LB}})^2 \times h_{\text{LB}}}\]
\[D_{\text{UB}} = \frac{785}{\pi \times 3.35^2 \times 8.45}\]
\[\pi \times 3.35^2 \times 8.45 = \pi \times 11.2225 \times 8.45 = 94.830125 \pi \approx 297.9178\]
\[D_{\text{UB}} = \frac{785}{297.9178} \approx 2.63495\text{ g/cm}^3\]
Correct to 3 significant figures, \(D_{\text{UB}} = 2.63\text{ g/cm}^3\).

(a)
B1 for 3.35 cao

(b)
B1 for \(m_{\text{UB}} = 785\) and \(h_{\text{LB}} = 8.45\) seen
M1 for \(\frac{m_{\text{UB}}}{\pi \times (r_{\text{LB}})^2 \times h_{\text{LB}}}\) with their bounds \(780 < m_{\text{UB}} \le 785\), \(3.35 \le r_{\text{LB}} < 3.4\), \(8.45 \le h_{\text{LB}} < 8.5\)
M1 for evaluation of denominator as \(297.9\dots\) or \(94.83\pi\)
A1 for 2.63 or 2.634 to 2.635 nfww

(a) In parallelogram \(OACB\), \(\vec{AC} = \vec{OB} = \mathbf{b}\).
Since \(AP : PC = 2 : 1\), \(\vec{AP} = \frac{2}{3}\vec{AC} = \frac{2}{3}\mathbf{b}\).
\[\vec{OP} = \vec{OA} + \vec{AP} = \mathbf{a} + \frac{2}{3}\mathbf{b}\]

(b) Since \(\vec{OX} = k\vec{OP} = k\left(\mathbf{a} + \frac{2}{3}\mathbf{b}\right) = k\mathbf{a} + \frac{2}{3}k\mathbf{b}\):
\[\vec{AX} = \vec{OX} - \vec{OA} = \left(k\mathbf{a} + \frac{2}{3}k\mathbf{b}\right) - \mathbf{a} = (k - 1)\mathbf{a} + \frac{2}{3}k\mathbf{b}\]

(c) \(Q\) is the midpoint of \(BC\), and \(\vec{BC} = \vec{OA} = \mathbf{a}\).
\[\vec{OQ} = \vec{OB} + \frac{1}{2}\vec{BC} = \mathbf{b} + \frac{1}{2}\mathbf{a}\]
\[\vec{AQ} = \vec{OQ} - \vec{OA} = \mathbf{b} + \frac{1}{2}\mathbf{a} - \mathbf{a} = -\frac{1}{2}\mathbf{a} + \mathbf{b}\]
Since \(A\), \(X\) and \(Q\) are collinear, \(\vec{AX} = \lambda \vec{AQ}\) for some scalar \(\lambda\):
\[(k - 1)\mathbf{a} + \frac{2}{3}k\mathbf{b} = \lambda\left(-\frac{1}{2}\mathbf{a} + \mathbf{b}\right) = -\frac{1}{2}\lambda\mathbf{a} + \lambda\mathbf{b}\]
Equating coefficients of \(\mathbf{b}\):
\[\lambda = \frac{2}{3}k\]
Equating coefficients of \(\mathbf{a}\):
\[k - 1 = -\frac{1}{2}\lambda = -\frac{1}{2}\left(\frac{2}{3}k\right) = -\frac{1}{3}k\]
\[k + \frac{1}{3}k = 1 \implies \frac{4}{3}k = 1 \implies k = \frac{3}{4}\]

(a)
M1 for \(\vec{AP} = \frac{2}{3}\mathbf{b}\) or valid route \(\vec{OA} + \frac{2}{3}\vec{OB}\)
A1 for \(\mathbf{a} + \frac{2}{3}\mathbf{b}\) oe

(b)
B1 for \((k - 1)\mathbf{a} + \frac{2}{3}k\mathbf{b}\) or \(k\mathbf{a} + \frac{2}{3}k\mathbf{b} - \mathbf{a}\)

(c)
M1 for \(\vec{AQ} = -\frac{1}{2}\mathbf{a} + \mathbf{b}\) oe and setting up ratio/equation with coefficients
A1 for \(k = \frac{3}{4}\) or 0.75

(a) Using the cosine rule in triangle \(ABC\):
\[BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC)\]
\[BC^2 = 55^2 + 70^2 - 2(55)(70)\cos(52^\circ)\]
\[BC^2 = 3025 + 4900 - 7700\cos(52^\circ)\]
\[BC^2 = 7925 - 7700(0.61566) = 7925 - 4740.59 = 3184.41\]
\[BC = \sqrt{3184.41} \approx 56.4305\text{ m}\]
Correct to 3 significant figures, \(BC = 56.4\text{ m}\).

(b) Since the pole \(BT\) is vertical and \(BC\) lies on the horizontal ground, triangle \(TBC\) is right-angled at \(B\).
Let \(\theta\) be the angle of elevation of \(T\) from \(C\):
\[\tan(\theta) = \frac{BT}{BC} = \frac{24}{56.4305} \approx 0.42530\]
\[\theta = \tan^{-1}(0.42530) \approx 23.039^\circ\]
Correct to 1 decimal place, the angle of elevation is \(23.0^\circ\).

(a)
M1 for \(55^2 + 70^2 - 2(55)(70)\cos(52^\circ)\)
A1 for 3184.4... seen
A1 for 56.4 or 56.43... nfww

(b)
M1 for \(\tan(\theta) = \frac{24}{\text{their } BC}\)
A1 for 23.0 or 23.03 to 23.04

(a)
\[\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 6 \\ 7 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 4 \\ 8 \end{pmatrix}\]

(b)
\[|\vec{AB}| = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} \approx 8.94427\]
Correct to 3 significant figures, \(|\vec{AB}| = 8.94\) (or exact \(4\sqrt{5}\)).

(c) The vector \(\vec{AC}\) is:
\[\vec{AC} = \vec{OC} - \vec{OA} = \begin{pmatrix} 10 \\ k \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 8 \\ k + 1 \end{pmatrix}\]
Since \(\angle BAC = 90^\circ\), \(\vec{AC}\) is perpendicular to \(\vec{AB}\).
The gradient of \(AB\) is \(m_{AB} = \frac{8}{4} = 2\).
Therefore, the gradient of \(AC\) is the negative reciprocal:
\[m_{AC} = -\frac{1}{2}\]
\[\frac{k + 1}{8} = -\frac{1}{2}\]
\[k + 1 = 8 \times \left(-\frac{1}{2}\right) = -4\]
\[k = -4 - 1 = -5\]

(a)
B1 for \(\begin{pmatrix} 4 \\ 8 \end{pmatrix}\)

(b)
M1 for \(\sqrt{4^2 + 8^2}\) or \(\sqrt{(\text{their } x)^2 + (\text{their } y)^2}\)
A1 for 8.94 or 8.944... or \(\sqrt{80}\) or \(4\sqrt{5}\)

(c)
M1 for \(\vec{AC} = \begin{pmatrix} 8 \\ k + 1 \end{pmatrix}\) and using perpendicular gradient \(m_{AC} = -\frac{1}{2}\) or scalar product \(4(8) + 8(k + 1) = 0\)
A1 for \(k = -5\)