2025 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

We can rewrite the negative fractional index as a positive fractional index by taking the reciprocal of the fraction:
\( \left(\frac{8}{27}\right)^{-\frac{2}{3}} = \left(\frac{27}{8}\right)^{\frac{2}{3}} \)

Next, apply the cube root first, then square the result:
\( \left(\frac{27}{8}\right)^{\frac{2}{3}} = \left(\sqrt[3]{\frac{27}{8}}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \) (or \( 2.25 \)).

M1 for \( \left(\frac{27}{8}\right)^{\frac{2}{3}} \) or showing cube root of \( \frac{8}{27} \) as \( \frac{2}{3} \) or showing square of reciprocal as \( \frac{729}{64} \)
A1 for \( \frac{9}{4} \) or \( 2\frac{1}{4} \) or \( 2.25 \)

First, factor out the common factor of 2:
\( 18x^2 - 50y^2 = 2(9x^2 - 25y^2) \)

Then, use the difference of two squares to factorise the expression inside the brackets:
\( 9x^2 - 25y^2 = (3x - 5y)(3x + 5y) \)

So, the completely factorised expression is:
\( 2(3x - 5y)(3x + 5y) \).

M1 for \( 2(9x^2 - 25y^2) \) or for a correct factorisation of their quadratic inside brackets, e.g. \( (3x - 5y)(6x + 10y) \)
A1 for \( 2(3x - 5y)(3x + 5y) \) or \( 2(3x + 5y)(3x - 5y) \)

Multiply every term by the common denominator 12 to clear the fractions:
\( 3(3x - 1) - 4(x + 2) = 24 \)

Expand the brackets:
\( 9x - 3 - 4x - 8 = 24 \)

Simplify the left-hand side:
\( 5x - 11 = 24 \)

Add 11 to both sides:
\( 5x = 35 \)

Divide by 5:
\( x = 7 \).

M1 for \( 3(3x - 1) - 4(x + 2) = 24 \) or for \( \frac{3(3x - 1) - 4(x + 2)}{12} = 2 \)
A1 for 7

We first find the number of students who play at least one of the sports using the principle of inclusion-exclusion:
\( n(F \cup T) = n(F) + n(T) - n(F \cap T) \)
\( n(F \cup T) = 18 + 15 - 5 = 28 \)

The number of students who play neither sport is the complement of this set within the universal set:
\( n((F \cup T)') = n(\mathscr{E}) - n(F \cup T) = 30 - 28 = 2 \).

M1 for \( 18 + 15 - 5 \) or for a Venn diagram with at least two correct regions correctly filled (e.g., 13, 5, 10)
A1 for 2

Express both numbers with the same power of 10:
\( 4.5 \times 10^7 - 0.8 \times 10^7 \)

Now, subtract the coefficients:
\( (4.5 - 0.8) \times 10^7 = 3.7 \times 10^7 \).

M1 for \( 45 \times 10^6 - 8 \times 10^6 \) or \( 4.5 \times 10^7 - 0.8 \times 10^7 \) or \( 37\,000\,000 \)
A1 for \( 3.7 \times 10^7 \)

First, find the gradient of the line passing through \( (2, -3) \) and \( (5, 6) \):
\( m_1 = \frac{6 - (-3)}{5 - 2} = \frac{9}{3} = 3 \)

The gradient of a perpendicular line is the negative reciprocal of \( m_1 \):
\( m_2 = -\frac{1}{m_1} = -\frac{1}{3} \).

M1 for gradient calculation: \( \frac{6 - (-3)}{5 - 2} \) or finding the gradient of the original line as 3
A1 for \( -\frac{1}{3} \)

The heaviest package corresponds to the ratio value of 9 parts.
So, 9 parts \( = 5.4\text{ kg} \).
This means 1 part \( = \frac{5.4}{9} = 0.6\text{ kg} \).

The total number of ratio parts is \( 2 + 5 + 9 = 16 \) parts.
Therefore, the total mass is:
\( 16 \times 0.6\text{ kg} = 9.6\text{ kg} \).

M1 for \( \frac{5.4}{9} \times (2 + 5 + 9) \) or for finding 0.6 or finding the three masses as 1.2, 3.0, and 5.4
A1 for 9.6

To find the lower bound of the perimeter, we need the lower bounds of the length and the width.

For length \( 65\text{ m} \) (nearest metre):
Lower bound of length \( = 64.5\text{ m} \).

For width \( 40\text{ m} \) (nearest \( 5\text{ m} \)):
Lower bound of width \( = 40 - 2.5 = 37.5\text{ m} \).

The lower bound for the perimeter is:
\( 2 \times (\text{length}_{\text{min}} + \text{width}_{\text{min}}) = 2 \times (64.5 + 37.5) = 2 \times 102 = 204\text{ m} \).

M1 for identification of both lower bounds: 64.5 and 37.5 (or for showing one of these bounds correctly)
A1 for 204

Multiply all terms by 12 to clear the fractions: \(3(2x - 3) - 4(x + 1) = 12\). Expand the brackets: \(6x - 9 - 4x - 4 = 12\). Simplify the equation: \(2x - 13 = 12\). Add 13 to both sides: \(2x = 25\). Divide by 2: \(x = 12.5\).

M1 for \(3(2x - 3) - 4(x + 1) = 12\) or better oe. A1 for 12.5 or \(\frac{25}{2}\).

Group the terms in pairs: \((6ax - 9ay) - (4bx - 6by)\). Factorise each pair: \(3a(2x - 3y) - 2b(2x - 3y)\). Factor out the common bracket: \((3a - 2b)(2x - 3y)\).

M1 for \(3a(2x - 3y) - 2b(2x - 3y)\) or \(2x(3a - 2b) - 3y(3a - 2b)\). A1 for \((3a - 2b)(2x - 3y)\) oe.

Multiply the coefficients: \(4 \times 5 = 20\). Multiply the powers of 10: \(10^{6} \times 10^{-11} = 10^{-5}\). Combine and convert to standard form: \(20 \times 10^{-5} = 2 \times 10^{-4}\).

M1 for \(20 \times 10^{-5}\) or \(0.0002\) oe. A1 for \(2 \times 10^{-4}\).

Let \(p\) be the original price of the jacket. An increase of 15% means the new price is 1.15 times the original price: \(1.15p = 92\). Solve for \(p\): \(p = \frac{92}{1.15} = \frac{9200}{115} = 80\).

M1 for \(\frac{92}{1.15}\) or \(92 \times \frac{100}{115}\) oe. A1 for 80.

First list the elements of each set within the universal set \(\mathscr{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\). \(A = \{2, 3, 5, 7\}\) and \(B = \{1, 3, 5, 7, 9\}\). The union is \(A \cup B = \{1, 2, 3, 5, 7, 9\}\). By De Morgan's laws, \(A' \cap B' = (A \cup B)'\). Therefore, \((A \cup B)' = \{4, 6, 8, 10\}\). The number of elements in this set is 4.

M1 for identifying \(A \cup B = \{1, 2, 3, 5, 7, 9\}\) or listing the elements of \(A' \cap B'\) with at most one error or omission. A1 for 4.

Calculate \(3\vec{p}\): \(3 \begin{pmatrix} 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 9 \\ -12 \end{pmatrix}\). Calculate \(2\vec{q}\): \(2 \begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}\). Subtract the two results: \(\begin{pmatrix} 9 \\ -12 \end{pmatrix} - \begin{pmatrix} -2 \\ 4 \end{pmatrix} = \begin{pmatrix} 9 - (-2) \\ -12 - 4 \end{pmatrix} = \begin{pmatrix} 11 \\ -16 \end{pmatrix}\).

M1 for \(\begin{pmatrix} 9 \\ -12 \end{pmatrix}\) or \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) seen or implied. A1 for \(\begin{pmatrix} 11 \\ -16 \end{pmatrix}\).

Use the volume formula for a cylinder, \(V = \pi r^2 h\), to find the radius \(r\): \(360\pi = \pi \times r^2 \times 10\). Simplifying gives \(360 = 10r^2 \implies r^2 = 36\), so the radius \(r = 6\text{ cm}\). Now calculate the curved surface area using \(A = 2\pi r h\): \(A = 2 \times \pi \times 6 \times 10 = 120\pi\text{ cm}^2\).

M1 for finding \(r = 6\) using the volume formula. A1 for \(120\pi\).

From the first equation, we can express \(y\) in terms of \(x\):
\(y = x + 2\)

Substitute this expression for \(y\) into the second equation:
\(x^2 + (x + 2)^2 = 10\)

Expand the brackets:
\(x^2 + x^2 + 4x + 4 = 10\)
\(2x^2 + 4x + 4 = 10\)

Subtract 10 from both sides to form a quadratic equation equal to zero:
\(2x^2 + 4x - 6 = 0\)

Divide the entire equation by 2:
\(x^2 + 2x - 3 = 0\)

Factorize the quadratic equation:
\((x + 3)(x - 1) = 0\)

This gives the two possible values for \(x\):
\(x = -3\) or \(x = 1\)

Now, substitute these back into the linear equation to find the corresponding \(y\) values:
When \(x = -3\):
\(y = -3 + 2 = -1\)

When \(x = 1\):
\(y = 1 + 2 = 3\)

So the solutions are \(x = -3, y = -1\) and \(x = 1, y = 3\).

**[M1]** for substitution of \(y = x + 2\) (or equivalent) into the quadratic equation.
**[A1]** for simplification to the correct three-term quadratic equation equal to zero (e.g., \(x^2 + 2x - 3 = 0\) or \(2x^2 + 4x - 6 = 0\)).
**[M1]** for factorizing or solving their quadratic equation to find two values of \(x\) (or \(y\)).
**[A1]** for both correct pairs of values: \(x = -3, y = -1\) and \(x = 1, y = 3\).

Using the Cosine Rule:
\(BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(BAC)\)

Substitute the given values into the formula:
\(BC^2 = 5^2 + 3^2 - 2 \times 5 \times 3 \times \cos(120^\circ)\)
\(BC^2 = 25 + 9 - 30 \times \cos(120^\circ)\)

Since \(\cos(120^\circ) = -\cos(60^\circ) = -0.5\):
\(BC^2 = 34 - 30 \times (-0.5)\)
\(BC^2 = 34 + 15\)
\(BC^2 = 49\)

Take the square root to find the length of \(BC\):
\(BC = \sqrt{49} = 7\text{ cm}\).

**[M1]** for correctly stating or using the Cosine Rule formula with values: \(BC^2 = 5^2 + 3^2 - 2 \times 5 \times 3 \times \cos(120^\circ)\).
**[B1]** for identifying that \(\cos(120^\circ) = -0.5\) (or equivalent).
**[M1]** for correct evaluation leading to \(BC^2 = 49\).
**[A1]** for the final answer \(7\).

First, factorize the numerator:
\(2x^2 + 5x - 3 = (2x - 1)(x + 3)\)

Next, factorize the denominator (using the difference of two squares):
\(x^2 - 9 = (x - 3)(x + 3)\)

Substitute the factorized expressions back into the fraction:
\(\frac{(2x - 1)(x + 3)}{(x - 3)(x + 3)}\)

Cancel the common factor \((x + 3)\) from both numerator and denominator:
\(\frac{2x - 1}{x - 3}\)

**[B1]** for correct factorization of the denominator: \((x - 3)(x + 3)\).
**[M2]** for correct factorization of the numerator: \((2x - 1)(x + 3)\) (or **[M1]** for a partial factorization of the form \((2x + a)(x + b)\) where \(ab = -3\) or \(2b + a = 5\)).
**[A1]** for the final answer \(\frac{2x - 1}{x - 3}\).

First, we need to find the slant height, \(l\), of the cone using Pythagoras' theorem:
\(l^2 = r^2 + h^2\)
\(l^2 = 5^2 + 12^2\)
\(l^2 = 25 + 144 = 169\)
\(l = \sqrt{169} = 13\text{ cm}\)

The formula for the total surface area (TSA) of a cone is:
\(\text{TSA} = \pi r^2 + \pi r l\)

Substitute the values \(r = 5\) and \(l = 13\) into the formula:
\(\text{TSA} = \pi (5)^2 + \pi (5)(13)\)
\(\text{TSA} = 25\pi + 65\pi = 90\pi\text{ cm}^2\)

**[M1]** for using Pythagoras' theorem to find the slant height: \(l^2 = 5^2 + 12^2\).
**[A1]** for finding the slant height \(l = 13\).
**[M1]** for substituting \(r = 5\) and their \(l\) into the total surface area formula \(\pi r^2 + \pi r l\).
**[A1]** for the final answer \(90\pi\).

First, find the critical values by factorizing the quadratic expression:
\(2x^2 - 5x - 3 = 0\)
\((2x + 1)(x - 3) = 0\)

This gives the critical values:
\(x = -0.5\) (or \(-\frac{1}{2}\)) and \(x = 3\)

Since the inequality is \(< 0\), we look for the region where the quadratic graph lies below the x-axis, which is between the two critical values:
\(-0.5 < x < 3\)

**[M1]** for a correct method to find the critical values, e.g., attempting to factorize \(2x^2 - 5x - 3\).
**[A1]** for obtaining the correct critical values: \(x = -0.5\) (or \(-\frac{1}{2}\)) and \(x = 3\).
**[M1]** for establishing the correct inequality structure (using 'between' rather than 'outside' regions).
**[A1]** for the correct final answer: \(-0.5 < x < 3\) (or \(-\frac{1}{2} < x < 3\)).

First, find the vector \(\vec{a} + 2\vec{b}\):
\(\vec{a} + 2\vec{b} = \begin{pmatrix} k \\ 6 \end{pmatrix} + 2\begin{pmatrix} 3 \\ -2 \end{pmatrix}\)
\(\vec{a} + 2\vec{b} = \begin{pmatrix} k \\ 6 \end{pmatrix} + \begin{pmatrix} 6 \\ -4 \end{pmatrix} = \begin{pmatrix} k + 6 \\ 2 \end{pmatrix}\)

The magnitude of this vector is given by:
\(|\vec{a} + 2\vec{b}| = \sqrt{(k + 6)^2 + 2^2}\)

We are given that this magnitude is \(\sqrt{85}\), so:
\(\sqrt{(k + 6)^2 + 4} = \sqrt{85}\)

Square both sides to remove the square root:
\((k + 6)^2 + 4 = 85\)
\((k + 6)^2 = 81\)

Take the square root of both sides:
\(k + 6 = 9\) or \(k + 6 = -9\)

Solve both equations:
\(k = 3\) or \(k = -15\).

**[M1]** for finding the combined vector \(\vec{a} + 2\vec{b} = \begin{pmatrix} k + 6 \\ 2 \end{pmatrix}\).
**[M1]** for setting up the magnitude equation: \((k + 6)^2 + 2^2 = 85\).
**[M1]** for simplifying the equation to \((k + 6)^2 = 81\) or \(k^2 + 12x - 45 = 0\).
**[A1]** for both correct values of \(k\): \(3\) and \(-15\).

Express both bases as powers of 2:
\(8 = 2^3\)
\(\frac{1}{4} = 2^{-2}\)

Substitute these into the equation:
\((2^3)^{x+1} = (2^{-2})^{2x-5}\)

Apply the laws of indices to simplify the powers:
\(2^{3(x+1)} = 2^{-2(2x-5)}\)
\(2^{3x+3} = 2^{-4x+10}\)

Since the bases are the same, we can equate the exponents:
\(3x + 3 = -4x + 10\)

Rearrange to solve for \(x\):
\(7x = 7\)
\(x = 1\).

**[M1]** for expressing LHS with base 2: \(2^{3(x+1)}\) (or equivalent).
**[M1]** for expressing RHS with base 2: \(2^{-2(2x-5)}\) (or equivalent).
**[M1]** for equating the indices: \(3x + 3 = -4x + 10\) (or equivalent linear equation from their powers).
**[A1]** for the final answer \(1\).

First, rearrange the equation to make \(\sin(x)\) the subject:
\(2\sin(x) = -\sqrt{3}\)
\(\sin(x) = -\frac{\sqrt{3}}{2}\)

The basic acute angle (or reference angle), \(\alpha\), where \(\sin(\alpha) = \frac{\sqrt{3}}{2}\) is:
\(\alpha = 60^\circ\)

Since \(\sin(x)\) is negative, \(x\) must lie in the 3rd and 4th quadrants.

In the 3rd quadrant:
\(x = 180^\circ + 60^\circ = 240^\circ\)

In the 4th quadrant:
\(x = 360^\circ - 60^\circ = 300^\circ\)

So the two solutions are \(240^\circ\) and \(300^\circ\).

**[M1]** for rearranging to \(\sin(x) = -\frac{\sqrt{3}}{2}\).
**[B1]** for identifying the reference angle of \(60^\circ\) (or finding one solution of \(240^\circ\) or \(300^\circ\)).
**[A1]** for the first correct solution: \(240^\circ\).
**[A1]** for the second correct solution: \(300^\circ\).

Using the formula for the area of a triangle:
\(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\text{angle } ABC)\)

\(12 = \frac{1}{2} \times 6 \times 5 \times \sin(\text{angle } ABC)\)

\(12 = 15\sin(\text{angle } ABC)\)

\(\sin(\text{angle } ABC) = \frac{12}{15} = \frac{4}{5}\)

Using the trigonometric identity \(\sin^2(\theta) + \cos^2(\theta) = 1\):
\(\cos^2(\text{angle } ABC) = 1 - \sin^2(\text{angle } ABC) = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}\)

\(\cos(\text{angle } ABC) = \pm\frac{3}{5}\)

Since angle \(ABC\) is obtuse (\(90^\circ < \text{angle } ABC < 180^\circ\)), the cosine of the angle must be negative.

Therefore, \(\cos(\text{angle } ABC) = -\frac{3}{5}\) (or \(-0.6\)).

M1 for \(12 = \frac{1}{2} \times 6 \times 5 \times \sin(\text{angle } ABC)\) oe
A1 for \(\sin(\text{angle } ABC) = \frac{4}{5}\) oe
M1 for \(\cos^2(\text{angle } ABC) = 1 - (\text{their } \sin(\text{angle } ABC))^2\)
A1 for \(-\frac{3}{5}\) or \(-0.6\) as the final answer

Factorise the numerator by grouping:
\(6ay - 9by + 8az - 12bz = 3y(2a - 3b) + 4z(2a - 3b)\)
\(= (2a - 3b)(3y + 4z)\)

Factorise the denominator as a difference of two squares:
\(9y^2 - 16z^2 = (3y - 4z)(3y + 4z)\)

Substitute the factorised expressions back into the fraction:
\(\frac{(2a - 3b)(3y + 4z)}{(3y - 4z)(3y + 4z)}\)

Cancel the common factor \((3y + 4z)\):
\(\frac{2a - 3b}{3y - 4z}\)

M1 for factorising the numerator by grouping, showing at least one correct step e.g. \(3y(2a - 3b)\) or \(4z(2a - 3b)\)
A1 for fully factorising the numerator: \((2a - 3b)(3y + 4z)\)
B1 for factorising the denominator: \((3y - 4z)(3y + 4z)\)
A1 for the correct final simplified fraction: \(\frac{2a - 3b}{3y - 4z}\)

Using the Cosine Rule: PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)cos(PQR). Substituting the given values: 7^2 = x^2 + (x + 3)^2 - 2x(x + 3)cos(60°). Since cos(60°) = 0.5, we have: 49 = x^2 + (x^2 + 6x + 9) - 2x(x + 3)(0.5) => 49 = x^2 + x^2 + 6x + 9 - x^2 - 3x => 49 = x^2 + 3x + 9 => x^2 + 3x - 40 = 0. Factorising this quadratic equation gives: (x + 8)(x - 5) = 0. This gives x = -8 or x = 5. Since x represents a length, it must be positive. Thus, x = 5.

M1 for correct substitution into cosine rule: 7^2 = x^2 + (x+3)^2 - 2x(x+3)cos(60°)
B1 for cos(60°) = 0.5 used
M1 for simplification to a three-term quadratic: x^2 + 3x - 40 = 0
M1 for factorising (x+8)(x-5) = 0 or using the quadratic formula correctly
A1 for x = 5 (discarding -8)

Substitute y = 2x - 3 into x^2 + y^2 = 26: x^2 + (2x - 3)^2 = 26 => x^2 + 4x^2 - 12x + 9 = 26 => 5x^2 - 12x - 17 = 0. Factorising the quadratic equation: (5x - 17)(x + 1) = 0. This gives x = 17/5 = 3.4 or x = -1. Substitute back to find y: For x = -1: y = 2(-1) - 3 = -5. For x = 3.4: y = 2(3.4) - 3 = 6.8 - 3 = 3.8. Thus, the solutions are x = -1, y = -5 and x = 3.4, y = 3.8.

M1 for substituting y = 2x - 3 into the quadratic equation
M1 for expanding and simplifying to 5x^2 - 12x - 17 = 0
M1 for solving their quadratic to find two values of x (e.g., factorising to (5x - 17)(x + 1) = 0)
A1 for x = -1, y = -5
A1 for x = 3.4 (or 17/5), y = 3.8 (or 19/5)

The total surface area of a solid hemisphere is the sum of its curved surface area and its flat circular base: Area_hemisphere = 2*pi*r^2 + pi*r^2 = 3*pi*r^2. The total surface area of a solid cylinder is the sum of its curved surface area and its two circular bases: Area_cylinder = 2*pi*r*h + 2*pi*r^2. Equating the two surface areas: 3*pi*r^2 = 2*pi*r*h + 2*pi*r^2. Subtract 2*pi*r^2 from both sides: pi*r^2 = 2*pi*r*h. Divide both sides by pi*r (since r is not 0): r = 2h => h = 0.5r.

M1 for total surface area of solid hemisphere as 3*pi*r^2
M1 for total surface area of solid cylinder as 2*pi*r*h + 2*pi*r^2
M1 for setting up the equation: 3*pi*r^2 = 2*pi*r*h + 2*pi*r^2
M1 for isolating h or simplifying the equation to pi*r^2 = 2*pi*r*h
A1 for h = 0.5r (or h = r/2)

Since OACB is a parallelogram, AC = OB = b and BC = OA = a. AP:PC = 2:1 means PC = 1/3 * AC = 1/3 * b. Since Q is the midpoint of BC, CQ = 1/2 * CB = -1/2 * BC = -1/2 * a. We can express the vector PQ as: PQ = PC + CQ = 1/3 * b - 1/2 * a.

M1 for identifying AC = b or BC = a
M1 for expressing PC = 1/3 * b (or AP = 2/3 * b)
M1 for expressing CQ = -1/2 * a (or BQ = 1/2 * a)
M1 for choosing a valid vector path, e.g., PQ = PC + CQ
A1 for 1/3 * b - 1/2 * a (or equivalent simplified form)

Factorise each quadratic expression: 1) 2x^2 + 5x - 3 = (2x - 1)(x + 3). 2) x^2 - 16 = (x - 4)(x + 4). 3) x^2 - 4x = x(x - 4). 4) 2x^2 - x = x(2x - 1). Substitute these factors back into the expression: ((2x - 1)(x + 3) / ((x - 4)(x + 4))) * (x(x - 4) / (x(2x - 1))). Cancel the common factors (2x - 1), (x - 4), and x from the numerator and denominator: (x + 3) / (x + 4).

M1 for factorising 2x^2 + 5x - 3 into (2x - 1)(x + 3)
M1 for factorising x^2 - 16 into (x - 4)(x + 4)
M1 for factorising x^2 - 4x into x(x - 4)
M1 for factorising 2x^2 - x into x(2x - 1)
A1 for (x + 3)/(x + 4)

Divide the equation by 2: x^2 - 4x + 1.5 = 0. Complete the square for x^2 - 4x: (x - 2)^2 - 4 + 1.5 = 0 => (x - 2)^2 - 2.5 = 0 => (x - 2)^2 = 2.5. Take the square root of both sides: x - 2 = +/- sqrt(2.5) => x - 2 = +/- sqrt(10/4) => x - 2 = +/- sqrt(10)/2. Add 2 to both sides: x = 2 +/- sqrt(10)/2. Thus, the exact solutions are x = 2 + sqrt(10)/2 and x = 2 - sqrt(10)/2.

M1 for dividing by 2 or starting to complete the square: 2(x^2 - 4x) + 3 = 0
M1 for completing the square inside: (x - 2)^2 - 4
M1 for rearranging to (x - 2)^2 = 2.5 (or 5/2)
M1 for taking square roots: x - 2 = +/- sqrt(10)/2
A1 for x = 2 + sqrt(10)/2 or x = 2 - sqrt(10)/2

Multiply all terms of the equation by 12 to clear the denominators:
\( 4x - 3(x-2) = 24 \)

Expand the brackets:
\( 4x - 3x + 6 = 24 \)

Simplify:
\( x + 6 = 24 \)

Subtract 6 from both sides:
\( x = 18 \)

M1 for multiplying through by 12 (or finding a correct common denominator, e.g. \( \frac{4x - 3(x-2)}{12} = 2 \))
A1 for 18

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

Divide both terms by \( 3a \):
\( 6a^2 \div 3a = 2a \)
\( 15ab \div 3a = 5b \)

Thus, the completely factorised expression is:
\( 3a(2a - 5b) \)

B1 for a partially factorised expression, e.g. \( 3(2a^2 - 5ab) \) or \( a(6a - 15b) \)
B2 for \( 3a(2a - 5b) \) final answer

Use the area formula for non-right-angled triangles:
\( \text{Area} = \frac{1}{2} a b \sin(C) \)

Substitute the given values:
\( \text{Area} = \frac{1}{2} \times 7 \times 10 \times \sin(150^\circ) \)

Calculate:
\( \text{Area} = 35 \times 0.5 = 17.5\text{ cm}^2 \)

M1 for \( \frac{1}{2} \times 7 \times 10 \times \sin(150) \) or equivalent
A1 for 17.5

Use the formula for the volume of a sphere:
\( V = \frac{4}{3} \pi r^3 \)

Substitute \( r = 4.5 \):
\( V = \frac{4}{3} \pi (4.5)^3 \)
\( V = \frac{4}{3} \pi \times 91.125 \)
\( V = 121.5\pi\text{ cm}^3 \)

M1 for \( \frac{4}{3} \times \pi \times 4.5^3 \) or better
A1 for \( 121.5\pi \) or \( \frac{243}{2}\pi \)

Multiply the coefficients and add the powers of 10:
\( (2.5 \times 6) \times 10^{4 + (-7)} = 15 \times 10^{-3} \)

Convert into standard form:
\( 1.5 \times 10^1 \times 10^{-3} = 1.5 \times 10^{-2} \)

B1 for \( 1.5 \times 10^{-2} \) or equivalent standard form representation

Let \( x \) be the original price of the coat.
Since it is reduced by \( 25\% \), the sale price is \( 75\% \) of \( x \):
\( 0.75x = 54 \)

Solve for \( x \):
\( x = \frac{54}{0.75} = 72 \)

M1 for \( 54 \div 0.75 \) or equivalent
A1 for 72

Solve the inequality:
\( 3n - 4 < 11 \)
\( 3n < 15 \)
\( n < 5 \)

The largest integer strictly less than 5 is 4.

B1 for 4

Rewrite the negative fractional power:
\( 16^{-\frac{3}{4}} = \frac{1}{16^{\frac{3}{4}}} \)

Calculate \( 16^{\frac{1}{4}} \):
\( 16^{\frac{1}{4}} = 2 \)

Cube the result:
\( 2^3 = 8 \)

So:
\( 16^{-\frac{3}{4}} = \frac{1}{8} \)

B1 for \( \frac{1}{8} \) or 0.125

Multiply the population by the average water consumption per person:
\((1.24 \times 10^6) \times (1.5 \times 10^2)\)
\(= (1.24 \times 1.5) \times (10^6 \times 10^2)\)
\(= 1.86 \times 10^8\)

M1 for \(1.24 \times 10^6 \times 1.5 \times 10^2\) or for the digits 186 seen
A1 for \(1.86 \times 10^8\)

Rewrite \(\frac{1}{27}\) as a power of 3:
\(\frac{1}{27} = \frac{1}{3^3} = 3^{-3}\)

Therefore,
\(3^{x - 2} = 3^{-3}\)

Equating the indices:
\(x - 2 = -3\)
\(x = -1\)

M1 for \(\frac{1}{27} = 3^{-3}\) or \(3^{x - 2} = 3^{-3}\) or \(3^x = \frac{1}{3}\) seen
A1 for \(-1\)

We use the compound interest formula: \(A = P\left(1 + \frac{r}{100}\right)^t\). Substituting the given values: \(9116.20 = 8000\left(1 + \frac{r}{100}\right)^4\). Dividing by 8000: \(\left(1 + \frac{r}{100}\right)^4 = 1.139525\). Taking the fourth root: \(1 + \frac{r}{100} = \sqrt[4]{1.139525} \approx 1.0332\). Therefore, \(\frac{r}{100} = 0.0332\) which gives \(r = 3.32\).

M1 for \(8000\left(1 + \frac{r}{100}\right)^4 = 9116.20\)
M1 for \(1 + \frac{r}{100} = \sqrt[4]{1.139525}\) or 1.0332 seen
A1 for 3.32

Using the Sine Rule: \(\frac{\sin(PRQ)}{PQ} = \frac{\sin(QPR)}{QR}\) which gives \(\frac{\sin(PRQ)}{15} = \frac{\sin(25^\circ)}{12}\). Solving for \(\sin(PRQ)\): \(\sin(PRQ) = \frac{15 \sin(25^\circ)}{12} \approx 0.52827\). The acute angle is \(PRQ \approx 31.9^\circ\). Since the angle is obtuse, \(PRQ = 180^\circ - 31.9^\circ = 148.1^\circ\).

M1 for \(\frac{\sin(PRQ)}{15} = \frac{\sin(25^\circ)}{12}\)
M1 for acute angle \(31.9^\circ\) or \(\sin(PRQ) \approx 0.528\)
A1 for 148.1

The total surface area of a cone is given by \(A = \pi r^2 + \pi r l\). Substituting the known values: \(90\pi = \pi (5)^2 + \pi (5) l\). Dividing the entire equation by \(\pi\) gives: \(90 = 25 + 5l\). Subtracting 25: \(65 = 5l\). Dividing by 5 gives: \(l = 13\text{ cm}\).

M1 for \(\pi (5)^2 + \pi (5) l = 90\pi\)
M1 for \(25 + 5l = 90\) or \(5l = 65\)
A1 for 13

Multiply both sides by \((5 - x)\): \(y(5 - x) = 3x + 2\). Expand the brackets: \(5y - xy = 3x + 2\). Collect terms with \(x\) on one side: \(5y - 2 = 3x + xy\). Factorise \(x\) on the right side: \(5y - 2 = x(3 + y)\). Divide by \((3 + y)\) to make \(x\) the subject: \(x = \frac{5y - 2}{y + 3}\).

M1 for \(y(5 - x) = 3x + 2\)
M1 for isolating \(x\) terms: \(5y - 2 = x(3 + y)\) or equivalent
A1 for \(x = \frac{5y - 2}{y + 3}\) (or equivalent)

We use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 3\), \(b = -8\), and \(c = -5\). Substituting these values: \(x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(-5)}}{2(3)}\) which simplifies to \(x = \frac{8 \pm \sqrt{64 + 60}}{6} = \frac{8 \pm \sqrt{124}}{6}\). Evaluating the two solutions: \(x_1 = \frac{8 + 11.1355}{6} \approx 3.19\) and \(x_2 = \frac{8 - 11.1355}{6} \approx -0.52\).

M1 for substitution into formula: \(\frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(-5)}}{2(3)}\) (allow one sign error)
B1 for \(\sqrt{124}\) or \(11.135...\) seen
A1 for -0.52 and 3.19

The area of a sector of a circle is given by the formula \(A = \frac{x}{360} \times \pi r^2\). Substituting the given values: \(36 = \frac{x}{360} \times \pi \times 9^2\) which simplifies to \(36 = \frac{81\pi x}{360}\). Rearranging to solve for \(x\): \(x = \frac{36 \times 360}{81\pi} = \frac{160}{\pi} \approx 50.9296^\circ\). Therefore, the sector angle correct to 1 decimal place is \(50.9^\circ\).

M1 for \(\frac{x}{360} \times \pi \times 9^2 = 36\) or equivalent
M1 for \(x = \frac{36 \times 360}{81\pi}\) or equivalent
A1 for 50.9 (accept 50.9 to 51.0)

First calculate the vector \(2\vec{a} + \vec{b}\): \(2\vec{a} = 2\begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} 10 \\ -4 \end{pmatrix}\). Now add \(\vec{b}\): \(2\vec{a} + \vec{b} = \begin{pmatrix} 10 \\ -4 \end{pmatrix} + \begin{pmatrix} -3 \\ 6 \end{pmatrix} = \begin{pmatrix} 7 \\ 2 \end{pmatrix}\). Next, find the magnitude of this resulting vector: \(|2\vec{a} + \vec{b}| = \sqrt{7^2 + 2^2} = \sqrt{49 + 4} = \sqrt{53} \approx 7.28\).

M1 for finding vector \(2\vec{a} + \vec{b} = \begin{pmatrix} 7 \\ 2 \end{pmatrix}\)
M1 for \(\sqrt{(\text{their } 7)^2 + (\text{their } 2)^2}\)
A1 for 7.28 (accept 7.28 to 7.2801)

There are 12 apples in total. The probability that both apples are red is: \(P(\text{Red, Red}) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132}\). The probability that both apples are green is: \(P(\text{Green, Green}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}\). The probability that both apples are the same colour is: \(P(\text{Same Colour}) = \frac{42}{132} + \frac{20}{132} = \frac{62}{132} = \frac{31}{66}\).

M1 for \(\frac{7}{12} \times \frac{6}{11}\) or \(\frac{5}{12} \times \frac{4}{11}\)
M1 for adding their two probabilities of the same-colour outcomes
A1 for \(\frac{31}{66}\) (or equivalent fraction)

Using the cosine rule:
\(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)\)
\(AC^2 = 7.2^2 + 5.4^2 - 2 \times 7.2 \times 5.4 \times \cos(112^\circ)\)
\(AC^2 = 51.84 + 29.16 - 77.76 \times (-0.374607)\)
\(AC^2 = 81 + 29.129\)
\(AC^2 = 110.129\)
\(AC = \sqrt{110.129} \approx 10.494\text{ cm}\)

M2 for \(7.2^2 + 5.4^2 - 2 \times 7.2 \times 5.4 \times \cos(112)\) or M1 for correct substitution of values into the cosine rule
A1 for 10.5 or 10.49 to 10.50

First, simplify the numerator:
\(5.4 \times 10^7 - 0.85 \times 10^7 = 4.55 \times 10^7\)
Now, divide by the denominator:
\(\frac{4.55 \times 10^7}{2.5 \times 10^{-3}} = 1.82 \times 10^{7 - (-3)} = 1.82 \times 10^{10}\)

M1 for conversion to same power of 10, e.g. \(54 \times 10^6 - 8.5 \times 10^6\) or \(4.55 \times 10^7\) seen
M1 for division of their numerator by \(2.5 \times 10^{-3}\)
A1 for \(1.82 \times 10^{10}\) as final answer

Volume of cylinder: \(V = \pi r^2 h = \pi \times 4.5^2 \times 12 = 243\pi\text{ cm}^3\)
Since the volume remains the same, volume of sphere = \(243\pi\text{ cm}^3\)
\(\frac{4}{3} \pi R^3 = 243\pi\)
\(\frac{4}{3} R^3 = 243\)
\(R^3 = 182.25\)
\(R = \sqrt[3]{182.25} \approx 5.67\text{ cm}\)

M1 for setting up \(\pi \times 4.5^2 \times 12\)
M1 for equating their volume to \(\frac{4}{3}\pi R^3\)
A1 for 5.67 or 5.669 to 5.670

Using the compound interest formula:
\(A = P(1 + r)^n\)
\(A = 150\,000 \times (1.08)^6\)
\(A = 150\,000 \times 1.5868743\)
\(A \approx 238\,031.15\)
Rounding to the nearest dollar gives \(238\,031\).

M2 for \(150\,000 \times 1.08^6\) or M1 for \(150\,000 \times 1.08^k\) where \(k > 1\)
A1 for 238031

Multiply all terms by the common denominator \((x-2)(x+1)\):
\(4(x+1) + 3(x-2) = 2(x-2)(x+1)\)
\(4x + 4 + 3x - 6 = 2(x^2 - x - 2)\)
\(7x - 2 = 2x^2 - 2x - 4\)
\(2x^2 - 9x - 2 = 0\)
Apply the quadratic formula:
\(x = \frac{9 \pm \sqrt{(-9)^2 - 4(2)(-2)}}{2(2)}\)
\(x = \frac{9 \pm \sqrt{81 + 16}}{4} = \frac{9 \pm \sqrt{97}}{4}\)
\(x_1 = \frac{9 + 9.848857}{4} \approx 4.71\)
\(x_2 = \frac{9 - 9.848857}{4} \approx -0.21\)

M1 for correctly eliminating fractions to obtain a 3-term quadratic, e.g., \(2x^2 - 9x - 2 = 0\)
M1 for substituting their coefficients correctly into the quadratic formula
A1 for both -0.21 and 4.71 (or answers rounding to these)

Total number of pens = 12.
Probability of choosing two red pens: \(P(R, R) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}\)
Probability of choosing two blue pens: \(P(B, B) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132}\)
Probability of same colour: \(P(\text{same}) = \frac{20}{132} + \frac{42}{132} = \frac{62}{132} = \frac{31}{66} \approx 0.470\)

M1 for \(\frac{5}{12} \times \frac{4}{11}\) or \(\frac{7}{12} \times \frac{6}{11}\)
M1 for adding their two products
A1 for \(\frac{31}{66}\) or \(0.470\) or \(0.4697\) to \(0.4700\)

Find the derivative of \(y\):
\(\frac{dy}{dx} = 4x - 12\)
Set the derivative to 0 to find the turning point:
\(4x - 12 = 0 \implies x = 3\)
Substitute \(x = 3\) back into the original equation:
\(y = 2(3)^2 - 12(3) + 7\)
\(y = 18 - 36 + 7 = -11\)
The turning point is \((3, -11)\).

M1 for finding the derivative \(4x - 12\) or completing the square to get \(2(x - 3)^2 - 11\)
M1 for setting \(4x - 12 = 0\) or identifying \(x = 3\)
A1 for \((3, -11)\)

Form the equation for the area:
\((2x + 1)(x - 2) = 33\)

Expand the brackets:
\(2x^2 - 4x + x - 2 = 33\)
\(2x^2 - 3x - 35 = 0\)

Factorise the quadratic:
\(2x^2 - 10x + 7x - 35 = 0\)
\(2x(x - 5) + 7(x - 5) = 0\)
\((2x + 7)(x - 5) = 0\)

So \(x = 5\) or \(x = -3.5\).
Since the width must be positive, \(x\) must be \(5\).

When \(x = 5\):
Length = \(2(5) + 1 = 11\text{ m}\)
Width = \(5 - 2 = 3\text{ m}\)

Perimeter = \(2 \times (11 + 3) = 28\text{ m}\).

M1 for setting up the equation \((2x + 1)(x - 2) = 33\)
M1 for expanding and simplifying to a three-term quadratic: \(2x^2 - 3x - 35 = 0\)
M1 for a correct method to solve their quadratic equation (e.g. factorisation \((2x+7)(x-5)=0\) or correct substitution into the quadratic formula)
A1 for \(x = 5\) (ignore negative root)
A1 for \(28\) as the final perimeter (cao)

Multiply through by the common denominator \((x-3)(x+2)\):
\(4(x+2) + 3(x-3) = 2(x-3)(x+2)\)

Expand both sides:
\(4x + 8 + 3x - 9 = 2(x^2 - x - 6)\)
\(7x - 1 = 2x^2 - 2x - 12\)

Rearrange into quadratic form:
\(2x^2 - 9x - 11 = 0\)

Factorise:
\((2x - 11)(x + 1) = 0\)

Solve:
\(2x - 11 = 0 \Rightarrow x = 5.5\)
\(x + 1 = 0 \Rightarrow x = -1\)

M1 for writing algebraic fractions with a common denominator (or multiplying through by the common denominator): \(4(x+2) + 3(x-3) = 2(x-3)(x+2)\)
M1 for expanding correctly: \(7x - 1 = 2(x^2 - x - 6)\) or \(2x^2 - 2x - 12\)
A1 for obtaining the correct quadratic equation: \(2x^2 - 9x - 11 = 0\)
M1 for solving their 3-term quadratic (factorising or quadratic formula)
A1 for both \(5.5\) and \(-1\) (or \(11/2\) and \(-1\)) (cao)

1. Factorise the numerator of the first fraction:
\(2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2)\)

2. Factorise the denominator of the first fraction:
\(x^2 + 5x + 6 = (x + 2)(x + 3)\)

3. Factorise the second fraction:
Numerator: \(2x - 4 = 2(x - 2)\)
Denominator: \(x^2 - 9 = (x - 3)(x + 3)\)

4. Rewrite the division as multiplication by the reciprocal:
$$\frac{2(x-2)(x+2)}{(x+2)(x+3)} \times \frac{(x-3)(x+3)}{2(x-2)}$$

5. Cancel common factors:
- \((x + 2)\) cancels from the first fraction.
- \((x - 2)\) cancels from numerator and denominator.
- \((x + 3)\) cancels from numerator and denominator.
- \(2\) cancels from numerator and denominator.

This leaves:
\(x - 3\)

B1 for factorising \(2x^2 - 8 = 2(x-2)(x+2)\)
B1 for factorising \(x^2 + 5x + 6 = (x+2)(x+3)\)
B1 for factorising \(x^2 - 9 = (x-3)(x+3)\)
M1 for multiplying by the reciprocal of the second fraction: \(\frac{\text{their } 2(x-2)(x+2)}{\text{their } (x+2)(x+3)} \times \frac{\text{their } (x-3)(x+3)}{\text{their } 2(x-2)}\)
A1 for \(x - 3\) as final answer (cao)

1. Square both sides to eliminate the square root:
\(q^2 = \frac{2p + 5}{3 - p}\)

2. Multiply both sides by \((3 - p)\):
\(q^2(3 - p) = 2p + 5\)

3. Expand the brackets:
\(3q^2 - p q^2 = 2p + 5\)

4. Move all terms containing \(p\) to one side and other terms to the other side:
\(3q^2 - 5 = 2p + p q^2\)

5. Factorise out \(p\):
\(3q^2 - 5 = p(2 + q^2)\)

6. Divide by \((2 + q^2)\) to make \(p\) the subject:
$$p = \frac{3q^2 - 5}{2 + q^2}$$
This can also be written as \(p = \frac{5 - 3q^2}{-2 - q^2}\).

M1 for squaring both sides: \(q^2 = \frac{2p + 5}{3 - p}\)
M1 for multiplying by denominator: \(q^2(3 - p) = 2p + 5\)
M1 for expanding and isolating \(p\) terms: \(3q^2 - 5 = 2p + p q^2\) (allow sign errors)
M1 for factorising out \(p\): \(3q^2 - 5 = p(2 + q^2)\)
A1 for \(p = \frac{3q^2 - 5}{2 + q^2}\) (or equivalent, e.g., \(p = \frac{5 - 3q^2}{-2 - q^2}\)) (cao)

1. Use the cosine rule to find the length of \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)
\(AC^2 = 7.2^2 + 5.4^2 - 2(7.2)(5.4)\cos(112^\circ)\)
\(AC^2 = 51.84 + 29.16 - 77.76 \times (-0.374606)\)
\(AC^2 = 81 + 29.13 = 110.13\)
\(AC = \sqrt{110.13} \approx 10.494\text{ cm}\)

2. Use the sine rule to find angle \(BAC\):
$$\frac{\sin(BAC)}{BC} = \frac{\sin(ABC)}{AC}$$
$$\frac{\sin(BAC)}{5.4} = \frac{\sin(112^\circ)}{10.494}$$
$$\sin(BAC) = \frac{5.4 \times \sin(112^\circ)}{10.494}$$
$$\sin(BAC) \approx \frac{5.4 \times 0.92718}{10.494} \approx 0.47711$$
Angle \(BAC = \sin^{-1}(0.47711) \approx 28.496^\circ\)
Correct to 1 decimal place, angle \(BAC\) is \(28.5^\circ\).

M1 for Cosine rule formula correctly applied: \(AC^2 = 7.2^2 + 5.4^2 - 2(7.2)(5.4)\cos(112^\circ)\)
A1 for \(AC = 10.49...\text{ or } 10.5\)
M1 for Sine rule formula correctly applied: \(\frac{\sin(BAC)}{5.4} = \frac{\sin(112^\circ)}{\text{their } AC}\)
M1 for rearranging to find \(\sin(BAC)\): \(\sin(BAC) = \frac{5.4 \times \sin(112^\circ)}{\text{their } AC}\)
A1 for \(28.5\) (accept \(28.49...\)) (cao)

1. Let the side lengths be \(a = 35\), \(b = 48\), and \(c = 62\). Use the Cosine rule to find one angle, for example, angle \(C\) opposite side \(c\):
$$c^2 = a^2 + b^2 - 2ab \cos(C)$$
$$62^2 = 35^2 + 48^2 - 2(35)(48) \cos(C)$$
$$3844 = 1225 + 2304 - 3360 \cos(C)$$
$$3844 = 3529 - 3360 \cos(C)$$
$$315 = -3360 \cos(C)$$
$$\cos(C) = -\frac{315}{3360} = -0.09375$$
$$C = \cos^{-1}(-0.09375) \approx 95.379^\circ$$

2. Use the area formula for a non-right-angled triangle:
$$\text{Area} = \frac{1}{2}ab \sin(C)$$
$$\text{Area} = \frac{1}{2}(35)(48) \sin(95.379^\circ)$$
$$\text{Area} = 840 \times 0.99560 \approx 836.30\text{ m}^2$$
Rounding to 3 significant figures gives \(836\text{ m}^2\).

M1 for Cosine rule formula correctly stated/applied: \(62^2 = 35^2 + 48^2 - 2(35)(48)\cos(C)\)
A1 for \(\cos(C) = -0.09375\) or \(C = 95.4^\circ\) (or equivalent for other angles: \(A = 34.2^\circ\), \(B = 50.4^\circ\))
M1 for using the area formula: \(\text{Area} = \frac{1}{2} \times 35 \times 48 \times \sin(95.4^\circ)\)
A1 for \(\sin(95.4^\circ) \approx 0.996\) or unrounded area \(836.3\)
A1 for \(836\) (cao)

1. Write down the formulas for the volume of the cylinder and the hemisphere:
Volume of cylinder = \(\pi r^2 h = \pi r^2 (3r) = 3\pi r^3\)
Volume of hemisphere = \(\frac{2}{3}\pi r^3\)

2. Express the total volume in terms of \(r\):
Total Volume = \(3\pi r^3 + \frac{2}{3}\pi r^3 = \frac{11}{3}\pi r^3\)

3. Set up the equation and solve for \(r\):
$$\frac{11}{3}\pi r^3 = 350$$
$$r^3 = \frac{350 \times 3}{11\pi}$$
$$r^3 = \frac{1050}{11\pi}$$
$$r^3 \approx 30.3837$$
$$r = \sqrt[3]{30.3837} \approx 3.1201$$
Rounding to 2 decimal places gives \(r = 3.12\).

M1 for volume of cylinder: \(\pi r^2 (3r)\) or \(3\pi r^3\)
M1 for volume of hemisphere: \(\frac{2}{3}\pi r^3\)
A1 for total volume equation: \(\frac{11}{3}\pi r^3 = 350\) (or \(3.67\pi r^3 = 350\))
M1 for rearranging to find \(r^3\) or \(r\): \(r = \sqrt[3]{\frac{1050}{11\pi}}\)
A1 for \(3.12\) (cao)

1. Use the total surface area formula for a cone to find the slant height \(l\):
$$\text{Total Surface Area} = \pi r^2 + \pi r l$$
$$32 = \pi (1.8)^2 + \pi (1.8) l$$
$$32 = 3.24\pi + 1.8\pi l$$
$$32 \approx 10.179 + 5.655 l$$
$$1.8\pi l = 32 - 3.24\pi \approx 21.821$$
$$l = \frac{21.821}{1.8\pi} \approx 3.859\text{ m}$$

2. Use Pythagoras' theorem in the right-angled triangle inside the cone to find the vertical height \(h\):
$$l^2 = r^2 + h^2$$
$$3.859^2 = 1.8^2 + h^2$$
$$14.891 = 3.24 + h^2$$
$$h^2 = 14.891 - 3.24 = 11.651$$
$$h = \sqrt{11.651} \approx 3.413\text{ m}$$
Rounding to 2 decimal places gives \(h = 3.41\text{ m}\).

M1 for total surface area formula correctly applied: \(\pi(1.8)^2 + \pi(1.8)l = 32\)
A1 for slant height \(l \approx 3.86\) (or unrounded \(3.8588...\))
M1 for Pythagoras' theorem relating \(h\), \(r\), and \(l\): \(h^2 = l^2 - 1.8^2\)
M1 for substituting their \(l\) into Pythagoras' theorem: \(h = \sqrt{\text{their } l^2 - 3.24}\)
A1 for \(3.41\) (accept \(3.41\) to \(3.42\)) (cao)