2025 Edexcel IGCSE Mathematics (Specification B) Practice Paper with Answers

Let the two consecutive odd numbers be \(2n-1\) and \(2n+1\), where \(n\) is an integer.

The difference between their squares is:
\((2n+1)^2 - (2n-1)^2\)
\(= (4n^2 + 4n + 1) - (4n^2 - 4n + 1)\)
\(= 4n^2 + 4n + 1 - 4n^2 + 4n - 1\)
\(= 8n\)

Since \(n\) is an integer, \(8n\) is always a multiple of 8.

Alternatively, letting the consecutive odd numbers be \(2n+1\) and \(2n+3\):
\((2n+3)^2 - (2n+1)^2\)
\(= (4n^2 + 12n + 9) - (4n^2 + 4n + 1)\)
\(= 8n + 8\)

\(= 8(n + 1)\), which is always a multiple of 8.

M1: Expressing two consecutive odd numbers algebraically, e.g., \(2n-1\) and \(2n+1\) (or \(2n+1\) and \(2n+3\)).
M1: Correctly expanding the squares of both expressions.
M1: Subtracting the two squared expressions to obtain a simplified expression, e.g., \(8n\) or \(8n+8\).
A1: Concluding with a clear explanation that the final expression is a multiple of 8 since \(n\) is an integer.

Since the particle is in equilibrium, the vector sum of all forces acting on it must equal zero:
\(\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}\)

\((2a\mathbf{i} - 5\mathbf{j}) + (3\mathbf{i} + b\mathbf{j}) + (-7\mathbf{i} + 2\mathbf{j}) = \mathbf{0}\)

Group the \(\mathbf{i}\) and \(\mathbf{j}\) components:
\((2a + 3 - 7)\mathbf{i} + (-5 + b + 2)\mathbf{j} = \mathbf{0}\)

\((2a - 4)\mathbf{i} + (b - 3)\mathbf{j} = \mathbf{0}\)

Equating the components to zero:
For \(\mathbf{i}\):
\(2a - 4 = 0 \implies a = 2\)

For \(\mathbf{j}\):
\(b - 3 = 0 \implies b = 3\)

M1: Setting the sum of the force vectors to the zero vector: \(\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}\).
M1: Grouping components and setting up at least one linear equation in \(a\) or \(b\).
A1: For finding \(a = 2\).
A1: For finding \(b = 3\).

Since \(\mathbf{M}\) is a singular matrix, its determinant must be equal to zero:
\(\det(\mathbf{M}) = (k+2)(k-2) - (3)(4) = 0\)

Expand the expression:
\(k^2 - 4 - 12 = 0\)

\(k^2 - 16 = 0\)

\(k^2 = 16\)

\(k = \pm 4\)

M1: Expressing the determinant of \(\mathbf{M}\) in terms of \(k\) and setting it to 0: \((k+2)(k-2) - (3)(4) = 0\).
M1: Expanding and simplifying to a quadratic equation: \(k^2 - 16 = 0\) (or equivalent).
A1: Obtaining \(k = 4\).
A1: Obtaining \(k = -4\) (or writing \(k = \pm 4\)).

The total surface area of a cylinder is given by:
\(A_{\text{cyl}} = 2\pi r^2 + 2\pi r h\)

Given that \(h = 3r\), substitute this into the formula:
\(A_{\text{cyl}} = 2\pi r^2 + 2\pi r(3r) = 2\pi r^2 + 6\pi r^2 = 8\pi r^2\)

The surface area of a sphere of radius \(R\) is given by:
\(A_{\text{sph}} = 4\pi R^2\)

Equating the two surface areas:
\(4\pi R^2 = 8\pi r^2\)

Divide both sides by \(4\pi\):
\(R^2 = 2r^2\)

Taking the square root (since radius must be positive):
\(R = r\sqrt{2}\)

M1: Writing down a correct expression for the total surface area of the cylinder in terms of \(r\), resulting in \(8\pi r^2\).
M1: Setting the surface area of the sphere equal to the total surface area of the cylinder: \(4\pi R^2 = 8\pi r^2\).
M1: Solving the equation for \(R^2\) to get \(R^2 = 2r^2\).
A1: Finding the correct simplified expression \(R = r\sqrt{2}\) (or \(R = \sqrt{2}r\)).

The rate of change of volume with respect to time is given by the derivative \(\frac{dV}{dt}\).

Differentiating \(V\) with respect to \(t\):
\(\frac{dV}{dt} = 12t^2 - 6t + 10\)

Now substitute \(t = 3\) into \(\frac{dV}{dt}\):
\(\frac{dV}{dt} = 12(3)^2 - 6(3) + 10\)

\(\frac{dV}{dt} = 12(9) - 18 + 10\)

\(\frac{dV}{dt} = 108 - 18 + 10 = 100\)

M1: Attempting to differentiate \(V\) with respect to \(t\) (at least one term differentiated correctly).
A1: Obtaining the fully correct derivative: \(\frac{dV}{dt} = 12t^2 - 6t + 10\).
M1: Substituting \(t = 3\) into their derivative expression.
A1: Obtaining the correct rate of change of \(100\).

The \(n\)-th term of a series, \(a_n\), is related to the sum of terms by the formula:
\(a_n = S_n - S_{n-1}\)

For the 10th term, we need to find:
\(a_{10} = S_{10} - S_9\)

First, calculate \(S_{10}\):
\(S_{10} = 2(10)^2 + 5(10) = 2(100) + 50 = 250\)

Next, calculate \(S_9\):
\(S_9 = 2(9)^2 + 5(9) = 2(81) + 45 = 162 + 45 = 207\)

Now, compute the 10th term:
\(a_{10} = 250 - 207 = 43\)

Alternatively, we can find the first few terms:
\(a_1 = S_1 = 2(1)^2 + 5(1) = 7\)

\(S_2 = 2(2)^2 + 5(2) = 18 \implies a_2 = S_2 - S_1 = 18 - 7 = 11\)

The common difference is \(d = a_2 - a_1 = 11 - 7 = 4\).

The 10th term is:
\(a_{10} = a_1 + 9d = 7 + 9(4) = 7 + 36 = 43\).

M1: For establishing a method to find the term using sums, e.g., \(a_{10} = S_{10} - S_9\) (or finding \(a_1 = 7\) and \(a_2 = 11\) to identify \(d = 4\)).
M1: Correctly evaluating \(S_{10} = 250\) (or evaluating \(a_1 = 7\) and \(a_2 = 11\)).
M1: Correctly evaluating \(S_9 = 207\) (or substituting their values into \(a_{10} = a_1 + 9d\)).
A1: Finding the correct 10th term of \(43\).

First, find the expression for the composite function \(\text{fg}(x)\):
\(\text{fg}(x) = \text{f}(\text{g}(x))\)

\(\text{fg}(x) = 2\left(\frac{3}{x+1}\right) - 5\)

\(\text{fg}(x) = \frac{6}{x+1} - 5\)

We are given that \(\text{fg}(x) = 1\):
\(\frac{6}{x+1} - 5 = 1\)

Add 5 to both sides:
\(\frac{6}{x+1} = 6\)

Multiply by \(x+1\):
\(6 = 6(x+1)\)

\(1 = x+1\)

\(x = 0\)

M1: Finding a correct expression for the composite function \(\text{fg}(x)\), e.g., \(2\left(\frac{3}{x+1}\right) - 5\).
M1: Setting their expression for \(\text{fg}(x)\) equal to 1.
M1: Correct algebraic manipulation to solve for \(x\), e.g., obtaining \(\frac{6}{x+1} = 6\) or \(6 = 6x + 6\).
A1: Finding the correct value \(x = 0\).

The total number of counters in the box is \(n + 5\).

The probability of selecting a red counter on the first draw is \(\frac{5}{n+5}\).

Since the selection is without replacement, the probability of selecting a red counter on the second draw is \(\frac{4}{n+4}\).

The probability that both counters are red is:

P(RR) = \frac{5}{n+5} \times \frac{4}{n+4} = \frac{20}{(n+5)(n+4)}

We are given that this probability is \(\frac{2}{7}\):

\frac{20}{(n+5)(n+4)} = \frac{2}{7}

Cross-multiplying yields:

2(n+5)(n+4) = 140

Divide by 2:

(n+5)(n+4) = 70


key expansions:

n^2 + 9n + 20 = 70


n^2 + 9n - 50 = 0

Factorising the quadratic equation:

(n + 14)(n - 5) = 0

This gives \(n = -14\) or \(n = 5\).

Since the number of blue counters \(n\) must be a positive integer, we reject \(n = -14\).

Therefore, \(n = 5\).

M1: Writing down a correct expression for the probability of selecting two red counters in terms of \(n\): \(\frac{5}{n+5} \times \frac{4}{n+4}\).
M1: Equating their probability expression to \(\frac{2}{7}\) and rearranging to form a quadratic equation of the form \(n^2 + 9n + c = 0\).
M1: Solving the quadratic equation \(n^2 + 9n - 50 = 0\) either by factorisation or by using the quadratic formula.
A1: Correct value \(n = 5\) (and rejecting the negative solution \(n = -14\)).

Let the three consecutive integers be \(n - 1\), \(n\), and \(n + 1\).

The sum of their squares is:
\((n - 1)^2 + n^2 + (n + 1)^2 = (n^2 - 2n + 1) + n^2 + (n^2 + 2n + 1)\)

Simplifying this expression:
\(= 3n^2 + 2\)

Since \(n\) is an integer, \(n^2\) is also an integer, which means \(3n^2\) is a multiple of 3.

Therefore, \(3n^2 + 2\) is always 2 more than a multiple of 3.

M1: Expressing three consecutive integers algebraically (e.g., \(n, n+1, n+2\) or \(n-1, n, n+1\)).
M1: Attempting to square and sum the three consecutive integers.
A1: Simplifying the algebraic expression to \(3n^2 + 2\) (or \(3(n^2 + 2n + 1) + 2\)).
A1: Completing the proof with a clear concluding statement linking the simplified expression to a multiple of 3 plus 2.

First, find the vector \(\overrightarrow{AB}\):
\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}\)

Since \(AP : PB = 3 : 1\), the vector \(\overrightarrow{AP}\) is:
\(\overrightarrow{AP} = \frac{3}{4}\overrightarrow{AB} = \frac{3}{4}(\mathbf{b} - \mathbf{a})\)

Now, find \(\overrightarrow{OP}\):
\(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \frac{3}{4}(\mathbf{b} - \mathbf{a}) = \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\)

Since \(OQ : QB = 2 : 1\), the vector \(\overrightarrow{OQ}\) is:
\(\overrightarrow{OQ} = \frac{2}{3}\mathbf{b}\)

Finally, find \(\overrightarrow{PQ}\):
\(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}\)
\(\overrightarrow{PQ} = \frac{2}{3}\mathbf{b} - \left(\frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\right)\)
\(\overrightarrow{PQ} = -\frac{1}{4}\mathbf{a} + \left(\frac{2}{3} - \frac{3}{4}\right)\mathbf{b}\)
\(\overrightarrow{PQ} = -\frac{1}{4}\mathbf{a} - \frac{1}{12}\mathbf{b}\)

M1: Finding \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) or \(\overrightarrow{AP} = \frac{3}{4}(\mathbf{b} - \mathbf{a})\).
M1: Finding a correct expression for \(\overrightarrow{OP} = \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\).
M1: Setting up the path \(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}\) (or equivalent path) with \(\overrightarrow{OQ} = \frac{2}{3}\mathbf{b}\).
A1: Correct simplified vector \(-\frac{1}{4}\mathbf{a} - \frac{1}{12}\mathbf{b}\) or equivalent single fraction form \(-\frac{1}{12}(3\mathbf{a} + \mathbf{b})\).

First, find the expression for velocity, \(v\), by differentiating the displacement \(s\) with respect to \(t\):
\(v = \frac{ds}{dt} = 3t^2 - 12t + 9\)

Next, find the expression for acceleration, \(a\), by differentiating the velocity \(v\) with respect to \(t\):
\(a = \frac{dv}{dt} = 6t - 12\)

We are given that the acceleration is \(12\text{ m/s}^2\):
\(6t - 12 = 12\)
\(6t = 24\)
\(t = 4\text{ seconds}\)

Now, substitute \(t = 4\) back into the expression for velocity:
\(v = 3(4)^2 - 12(4) + 9\)
\(v = 3(16) - 48 + 9\)
\(v = 48 - 48 + 9 = 9\text{ m/s}\)

M1: Differentiating \(s\) to find velocity: \(v = 3t^2 - 12t + 9\) (allow one slip).
M1: Differentiating their velocity expression to find acceleration: \(a = 6t - 12\).
M1: Equating their acceleration to 12 and solving to find \(t = 4\).
A1: Substituting \(t = 4\) into their velocity expression to obtain \(9\text{ m/s}\).

Using the trigonometric identity \(\sin^2 x = 1 - \cos^2 x\):
\(6(1 - \cos^2 x) - \cos x - 5 = 0\)
\(6 - 6\cos^2 x - \cos x - 5 = 0\)
\(-6\cos^2 x - \cos x + 1 = 0\)
\(6\cos^2 x + \cos x - 1 = 0\)

Let \(y = \cos x\):
\(6y^2 + y - 1 = 0\)
\((3y - 1)(2y + 1) = 0\)

This gives:
\(y = \frac{1}{3}\) or \(y = -\frac{1}{2}\)

Case 1: \(\cos x = \frac{1}{3}\)
\(x = \arccos\left(\frac{1}{3}\right) \approx 70.528^\circ \approx 70.5^\circ\)

Case 2: \(\cos x = -\frac{1}{2}\)
\(x = \arccos\left(-\frac{1}{2}\right) = 120^\circ\)

Both values lie within the interval \(0^\circ \le x \le 180^\circ\).

M1: Using \(\sin^2 x = 1 - \cos^2 x\) to construct a quadratic equation in terms of \(\cos x\).
M1: Solving the quadratic equation \(6\cos^2 x + \cos x - 1 = 0\) to obtain \(\cos x = \frac{1}{3}\) and \(\cos x = -\frac{1}{2}\).
A1: Finding one correct angle (either \(120^\circ\) or \(70.5^\circ\)).
A1: Finding both correct angles: \(70.5^\circ\) and \(120^\circ\) with no other incorrect solutions in the range.

The perimeter, \(P\), of a sector with radius \(r\) and arc length \(l\) is:
\(P = 2r + l\)

Since the perimeter is \(40\) cm:
\(2r + l = 40 \implies l = 40 - 2r\)

The formula for the area, \(A\), of a sector is:
\(A = \frac{1}{2} r l\)

Substituting \(l = 40 - 2r\) into the area formula:
\(A = \frac{1}{2} r (40 - 2r)\)
\(A = 20r - r^2\) (as required).

M1: Setting up the equation for the perimeter of the sector: \(2r + l = 40\) or \(2r + r\theta = 40\).
M1: Expressing the arc length \(l\) (or \(r\theta\)) in terms of \(r\): \(l = 40 - 2r\).
M1: Writing the correct formula for the area of a sector: \(A = \t\frac{1}{2}rl\) or \(A = \frac{1}{2}r^2\theta\).
A1: Substituting and completing the algebraic proof to arrive at \(A = 20r - r^2\) with no errors shown.

Method 1: Using triangle \(ADT\)
Since \(AB\) is a diameter, the angle in a semicircle is \(90^\circ\):
\(\angle ADB = 90^\circ\)

By the Alternate Segment Theorem, the angle between the tangent \(TD\) and the chord \(BD\) is equal to the angle in the alternate segment:
\(\angle BDT = \angle DAB = 28^\circ\)

Thus, the total angle \(\angle ADT\) is:
\(\angle ADT = \angle ADB + \angle BDT = 90^\circ + 28^\circ = 118^\circ\)

In triangle \(ADT\), the sum of angles is \(180^\circ\):
\(\angle ATD = 180^\circ - \angle DAT - \angle ADT\)
\(\angle ATD = 180^\circ - 28^\circ - 118^\circ = 34^\circ\)

Method 2: Using triangle \(ODT\)
Since \(OA\) and \(OD\) are both radii of the circle, triangle \(OAD\) is isosceles with \(OA = OD\):
\(\angle ODA = \angle OAD = 28^\circ\)
\(\angle AOD = 180^\circ - 2(28^\circ) = 124^\circ\)

Since \(ABT\) is a straight line:
\(\angle DOT = 180^\circ - 124^\circ = 56^\circ\)

The tangent \(TD\) is perpendicular to the radius \(OD\):
\(\angle ODT = 90^\circ\)

In right-angled triangle \(ODT\):
\(\angle ATD = 180^\circ - 90^\circ - 56^\circ = 34^\circ\)

M1: Recognizing angle \(\angle ADB = 90^\circ\) or calculating \(\angle AOD = 124^\circ\).
M1: Applying the alternate segment theorem to find \(\angle BDT = 28^\circ\) or finding \(\angle DOT = 56^\circ\).
M1: Setting up a complete angle sum calculation for \(\triangle ADT\) or \(\triangle ODT\).
A1: Correct final answer: \(34^\circ\) (or simply \(34\)).

The total number of sweets in the bag is \(5 + n\).

The probability of selecting a red sweet first is:
\(P(R_1) = \t\frac{5}{5+n}\)

After removing one red sweet, there are \(4\) red sweets left, and the total number of sweets remaining is \(4 + n\). The probability of selecting a second red sweet is:
\(P(R_2 | R_1) = \frac{4}{4+n}\)

The probability that both sweets are red is:
\(\frac{5}{5+n} \times \frac{4}{4+n} = \frac{2}{9}\)

Simplify the equation:
\(\frac{20}{(5+n)(4+n)} = \frac{2}{9}\)

Multiply both sides by 9 and divide by 2:
\(90 = (5+n)(4+n)\)
\(90 = 20 + 9n + n^2\)
\(n^2 + 9n - 70 = 0\)

Factor the quadratic equation:
\((n + 14)(n - 5) = 0\)

Since the number of green sweets \(n\) must be positive, we reject \(n = -14\).
Thus, \(n = 5\).

M1: Writing down the probability equation in terms of \(n\): \(\frac{5}{5+n} \times \frac{4}{4+n} = \frac{2}{9}\).
M1: Expanding the brackets and rearranging to form a quadratic equation of the form \(n^2 + 9n - 70 = 0\).
M1: Factorising or using the quadratic formula on their quadratic equation to obtain solutions.
A1: Concluding with \(n = 5\) only.

First, evaluate \(\text{g}(4)\):
\(\text{g}(4) = 3(4) - 2 = 12 - 2 = 10\)

We need to solve:
\(\text{f}^{-1}(x) = 10\)

By definition of inverse functions, this is equivalent to finding \(x\) such that:
\(x = \text{f}(10)\)

Substitute \(10\) into the definition of \(\text{f}(x)\):
\(x = \frac{2(10) + 1}{10 - 3} = \frac{21}{7} = 3\)

(Alternatively, finding \(\text{f}^{-1}(x)\) explicitly:
Let \(y = \frac{2x+1}{x-3} \implies yx - 3y = 2x + 1 \implies x(y-2) = 3y+1 \implies \text{f}^{-1}(x) = \frac{3x+1}{x-2}\).
Solving \(\frac{3x+1}{x-2} = 10 \implies 3x + 1 = 10x - 20 \implies 21 = 7x \implies x = 3\).)

M1: Finding \(\text{g}(4) = 10\).
M1: Understanding the relationship \(\text{f}^{-1}(x) = 10 \iff x = \text{f}(10)\) OR successfully finding the inverse function \(\text{f}^{-1}(x) = \frac{3x+1}{x-2}\).
M1: Evaluating \(\text{f}(10)\) OR setting up the equation \(\frac{3x+1}{x-2} = 10\) and multiplying out the denominator.
A1: Correctly obtaining \(x = 3\).

To find the velocity \(v\), we differentiate the displacement \(s\) with respect to \(t\):

\(v = \frac{ds}{dt} = 6t^2 - 18t + 12\)

The particle is momentarily at rest when \(v = 0\):

\(6t^2 - 18t + 12 = 0 \implies 6(t^2 - 3t + 2) = 0 \implies 6(t-1)(t-2) = 0\)

Thus, the particle is at rest at \(t = 1\) and \(t = 2\).

To find the acceleration \(a\), we differentiate the velocity \(v\) with respect to \(t\):

\(a = \frac{dv}{dt} = 12t - 18\)

When \(t = 1\):

\(a = 12(1) - 18 = -6\text{ m/s}^2\)

When \(t = 2\):

\(a = 12(2) - 18 = 6\text{ m/s}^2\)

Therefore, the values of the acceleration when momentarily at rest are \(-6\) and \(6\).

M1: Differentiates \(s\) with respect to \(t\) to find an expression for \(v\) (at least two terms correct).
M1: Sets their expression for \(v = 0\) and solves for \(t\) to find \(t = 1\) and \(t = 2\).
M1: Differentiates \(v\) to find an expression for \(a\) and substitutes at least one of their values of \(t\).
A1: Correct values of acceleration, \(-6\) and \(6\) (or \(\pm 6\)).

Let the two consecutive odd integers be \(2n - 1\) and \(2n + 1\), where \(n\) is an integer.

The sum of their squares is given by:

\(S = (2n - 1)^2 + (2n + 1)^2\)

Expanding both expressions:

\((2n - 1)^2 = 4n^2 - 4n + 1\)

\((2n + 1)^2 = 4n^2 + 4n + 1\)

Adding these together:

\(S = (4n^2 - 4n + 1) + (4n^2 + 4n + 1) = 8n^2 + 2\)

Since \(8n^2\) is a multiple of 8 for any integer \(n\), the expression \(8n^2 + 2\) is always 2 more than a multiple of 8.

M1: Defines two consecutive odd integers algebraically, e.g., \(2n-1\) and \(2n+1\) (or \(2n+1\) and \(2n+3\)).
M1: Sets up the sum of their squares, e.g., \((2n-1)^2 + (2n+1)^2\).
A1: Expands and simplifies correctly to \(8n^2 + 2\) (or \(8n^2 + 16n + 10\)).
C1: Provides a clear conclusion linking the simplified expression to the final proof (e.g., explaining that \(8n^2\) is a multiple of 8, so \(8n^2+2\) is 2 more than a multiple of 8).

Using Newton's Second Law, \(\mathbf{F} = m\mathbf{a}\):

\(3\mathbf{i} - 2\mathbf{j} = 2\mathbf{a} \implies \mathbf{a} = 1.5\mathbf{i} - \mathbf{j}\text{ m/s}^2\)

Since the particle starts from rest, its velocity \(\mathbf{v}\) after 4 seconds is:

\(\mathbf{v} = \mathbf{u} + \mathbf{a}t = \mathbf{0} + (1.5\mathbf{i} - \mathbf{j}) \times 4 = 6\mathbf{i} - 4\mathbf{j}\text{ m/s}\)

The speed of the particle is the magnitude of the velocity vector:

\(\text{Speed} = |\mathbf{v}| = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}\text{ m/s}\).

M1: Finds the acceleration vector \(\mathbf{a} = 1.5\mathbf{i} - \mathbf{j}\) using \(\mathbf{F} = m\mathbf{a}\).
M1: Integrates or uses \(\mathbf{v} = \mathbf{a}t\) to find the velocity vector \(\mathbf{v} = 6\mathbf{i} - 4\mathbf{j}\) at \(t=4\).
M1: Applies Pythagoras' theorem to find the magnitude of their velocity vector.
A1: Correct exact speed of \(2\sqrt{13}\) (or \(\sqrt{52}\)).

Using the determinant property \(\det(\mathbf{AB}) = \det(\mathbf{A}) \times \det(\mathbf{B})\):

First, find the determinant of \(\mathbf{A}\):

\(\det(\mathbf{A}) = (2)(3) - (k)(-1) = 6 + k\)

Next, find the determinant of \(\mathbf{B}\):

\(\det(\mathbf{B}) = (1)(5) - (-2)(4) = 5 + 8 = 13\)

Now set up the equation for \(\det(\mathbf{AB})\):

\(\det(\mathbf{AB}) = 13(6 + k) = 39\)

Divide by 13:

\(6 + k = 3 \implies k = -3\).

M1: Attempts to find \(\det(\mathbf{A})\) in terms of \(k\) or attempts to multiply the matrices to find \(\mathbf{AB}\).
M1: Obtains a correct expression for the determinant of \(\mathbf{A}\) as \(6+k\) or correctly multiplies \(\mathbf{AB} = \begin{pmatrix} 2+4k & -4+5k \\ 11 & 17 \end{pmatrix}\).
M1: Sets up the equation \(13(6+k) = 39\) or \(17(2+4k) - 11(-4+5k) = 39\) and solves for \(k\).
A1: Correct value of \(k = -3\).

First, find the composite function \(\mathrm{fg}(x)\):

\(\mathrm{fg}(x) = \mathrm{f}\left(\frac{2}{3-x}\right) = \frac{2}{3-x} + 1 = \frac{2 + (3-x)}{3-x} = \frac{5-x}{3-x}\)

Next, find the composite function \(\mathrm{gf}(x)\):

\(\mathrm{gf}(x) = \mathrm{g}(x+1) = \frac{2}{3-(x+1)} = \frac{2}{2-x}\)

Set \(\mathrm{fg}(x) = \mathrm{gf}(x)\):

\(\frac{5-x}{3-x} = \frac{2}{2-x}\)

Cross-multiply to clear denominators:

\((5-x)(2-x) = 2(3-x)\)

\(10 - 7x + x^2 = 6 - 2x\)

Rearrange into standard quadratic form:

\(x^2 - 5x + 4 = 0\)

Factor the quadratic:

\((x-1)(x-4) = 0\)

Thus, \(x = 1\) or \(x = 4\).

M1: Correctly substitutes to find a simplified expression for \(\mathrm{fg}(x) = \frac{5-x}{3-x}\) or \(\mathrm{gf}(x) = \frac{2}{2-x}\).
M1: Equates both composite expressions and removes denominators to form a quadratic equation.
M1: Standardizes the quadratic equation to \(x^2 - 5x + 4 = 0\) and attempts to solve by factoring or the formula.
A1: Both correct values \(x = 1\) and \(x = 4\).

(a) The total surface area \(A\) of a solid cylinder is:

\(A = 2\pi r^2 + 2\pi r h = 120\pi\)

Divide the entire equation by \(2\pi\):

\(r^2 + rh = 60 \implies rh = 60 - r^2 \implies h = \frac{60 - r^2}{r}\)

The volume \(V\) of the cylinder is:

\(V = \pi r^2 h = \pi r^2 \left(\frac{60 - r^2}{r}\right) = \pi r(60 - r^2) = 60\pi r - \pi r^3\)

(b) To maximize the volume, we differentiate \(V\) with respect to \(r\):

\(\frac{dV}{dr} = 60\pi - 3\pi r^2\)

Set \(\frac{dV}{dr} = 0\):

\(60\pi - 3\pi r^2 = 0 \implies 3\pi r^2 = 60\pi \implies r^2 = 20\)

Since \(r > 0\), we take the positive square root:

\(r = \sqrt{20} = 2\sqrt{5}\text{ cm}\).

M1: Expresses the surface area formula to make \(h\) the subject: \(h = \frac{60 - r^2}{r}\).
A1: Shows the algebraic steps to substitute \(h\) into \(V = \pi r^2 h\) to obtain \(V = 60\pi r - \pi r^3\).
M1: Differentiates \(V\) to find \(\frac{dV}{dr}\) and sets it equal to 0.
A1: Correct answer \(2\sqrt{5}\).

The total number of balls in the box is \(n + 5\).

The probability of drawing a red ball first is \(\frac{5}{n+5}\).

Since the drawing is without replacement, the probability of drawing a red ball second is \(\frac{4}{n+4}\).

The probability that both balls are red is:

\(\mathrm{P}(\text{Red, Red}) = \frac{5}{n+5} \times \frac{4}{n+4} = \frac{20}{(n+5)(n+4)}
\)

We are given this probability is \(\frac{2}{9}\):

\(\frac{20}{(n+5)(n+4)} = \frac{2}{9}\)

Cross-multiplying gives:

\(180 = 2(n+5)(n+4) \implies 90 = n^2 + 9n + 20\)

Rearranging into standard quadratic form:

\(n^2 + 9n - 70 = 0\)

Factorising the quadratic equation:

\((n + 14)(n - 5) = 0\)

Since \(n\) represents the number of green balls, it must be positive. Therefore, we reject \(n = -14\), leaving \(n = 5\).

M1: Formulates the correct probability for both balls being red: \(\frac{5}{n+5} \times \frac{4}{n+4}\).
M1: Sets the probability equal to \(\frac{2}{9}\) and expands to form a quadratic equation.
M1: Solves the quadratic equation \(n^2 + 9n - 70 = 0\) by factorisation or formula.
A1: Correct positive integer solution \(n = 5\).

Let \(x\) be the number of students who study all three subjects, i.e., \(|P \cap C \cap B| = x\).

The total number of students studying at least one of the three subjects is:

\(|P \cup C \cup B| = 40 - 4 = 36\)

Using the principle of inclusion-exclusion:

\(|P \cup C \cup B| = |P| + |C| + |B| - |P \cap C| - |P \cap B| - |C \cap B| + |P \cap C \cap B|\)

Substitute the known values into the equation:

\(36 = 23 + 18 + 17 - 9 - 8 - 7 + x\)

\(36 = 58 - 24 + x \implies 36 = 34 + x \implies x = 2\)

Now we determine the number of students studying at least two subjects, which corresponds to the regions: \((P \cap C) \cup (P \cap B) \cup (C \cap B)\):

- Students studying Physics and Chemistry only = \(9 - x = 9 - 2 = 7\)
- Students studying Physics and Biology only = \(8 - x = 8 - 2 = 6\)
- Students studying Chemistry and Biology only = \(7 - x = 7 - 2 = 5\)
- Students studying all three subjects = \(x = 2\)

Sum of students studying at least two subjects:

\(7 + 6 + 5 + 2 = 20\)

The probability that a student chosen at random studies at least two of the three subjects is:

\(\mathrm{P}(\text{At least 2}) = \frac{20}{40} = \frac{1}{2} = 0.5\).

M1: Applies inclusion-exclusion principle or sets up a correct Venn diagram to determine the number of students studying all three subjects, \(x = 2\).
M1: Identifies regions corresponding to at least two subjects: \((9-x) + (8-x) + (7-x) + x\).
M1: Correctly evaluates the sum of these regions to get 20 students.
A1: Correct final probability of \(\frac{1}{2}\) (or \(0.5\), or \(\frac{20}{40}\)).

The particle is instantaneously at rest when its velocity \(v = 0\).

\(3t^2 - 12t + 9 = 0\)

Dividing by 3:
\(t^2 - 4t + 3 = 0\)
\((t - 1)(t - 3) = 0\)

So, \(P\) is at rest when \(t = 1\) and \(t = 3\). The first time \(P\) is at rest is at \(t = 1\) second.

The acceleration \(a\) of \(P\) is given by the derivative of the velocity with respect to time \(t\):
\(a = \frac{dv}{dt} = 6t - 12\)

At \(t = 1\):
\(a = 6(1) - 12 = -6\text{ m/s}^2\)

M1: Sets \(v = 0\) and attempts to solve the quadratic equation \(3t^2 - 12t + 9 = 0\) to find at least one value of \(t\).
A1: Correctly identifies \(t = 1\) as the first time the particle is at rest.
M1: Differentiates the velocity expression to obtain \(a = 6t - 12\).
A1: Substitutes \(t = 1\) into their acceleration expression to get \(-6\).

Let two consecutive odd integers be represented as \(2n + 1\) and \(2n + 3\), where \(n\) is an integer.

The difference between their squares is:
\((2n + 3)^2 - (2n + 1)^2\)

Expanding both terms:
\((2n + 3)^2 = 4n^2 + 12n + 9\)
\((2n + 1)^2 = 4n^2 + 4n + 1\)

Subtracting the second expression from the first:
\((4n^2 + 12n + 9) - (4n^2 + 4n + 1) = 8n + 8\)

Factorising out 8:
\(8(n + 1)\)

Since \(n\) is an integer, \(n + 1\) is also an integer. Therefore, \(8(n + 1)\) is always a multiple of 8.

Alternative Method:
Let the consecutive odd integers be \(2n - 1\) and \(2n + 1\).
\((2n + 1)^2 - (2n - 1)^2 = (4n^2 + 4n + 1) - (4n^2 - 4n + 1) = 8n\).
Since \(n\) is an integer, \(8n\) is a multiple of 8.

M1: Expresses two consecutive odd integers algebraically, e.g., \(2n + 1\) and \(2n + 3\) (or \(2n - 1\) and \(2n + 1\)).
M1: Writes down the difference of their squares and attempts to expand both bracketed terms.
A1: Correctly simplifies the expanded expression to \(8n + 8\) (or \(8n\)).
C1: Factorises out 8 to show \(8(n+1)\) (or states that \(8n\) is a multiple of 8 since \(n\) is an integer) and concludes the proof.

The resultant force \(\mathbf{R}\) is the vector sum of the individual forces:
\(\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3\)
\(\mathbf{R} = (2a\mathbf{i} + 3\mathbf{j}) + (5\mathbf{i} - b\mathbf{j}) + (-3\mathbf{i} + 7\mathbf{j})\)
\(\mathbf{R} = (2a + 5 - 3)\mathbf{i} + (3 - b + 7)\mathbf{j}\)
\(\mathbf{R} = (2a + 2)\mathbf{i} + (10 - b)\mathbf{j}\)

According to Newton's second law, \(\mathbf{R} = m\mathbf{a}\), where \(m = 4\text{ kg}\) and \(\mathbf{a} = 2\mathbf{i} + 2\mathbf{j}\):
\(\mathbf{R} = 4(2\mathbf{i} + 2\mathbf{j}) = 8\mathbf{i} + 8\mathbf{j}\)

Equating the components:
For the \(\mathbf{i}\) component:
\(2a + 2 = 8 \implies 2a = 6 \implies a = 3\)

For the \(\mathbf{j}\) component:
\(10 - b = 8 \implies b = 2\)

M1: Sums the three forces to write an expression for the resultant force in terms of \(a\) and \(b\): \((2a + 2)\mathbf{i} + (10 - b)\mathbf{j}\).
M1: Applies \(\mathbf{F} = m\mathbf{a}\) to find the target resultant force vector: \(8\mathbf{i} + 8\mathbf{j}\).
A1: Equates components to obtain two linear equations: \(2a + 2 = 8\) and \(10 - b = 8\).
A1: Correctly solves both equations to find \(a = 3\) and \(b = 2\).

(a) Combine the fractions over a common denominator:

\(\frac{(x+5)x - 4(x-1)}{x(x-1)}\)

\(= \frac{x^2 + 5x - 4x + 4}{x(x-1)}\)

\(= \frac{x^2 + x + 4}{x(x-1)}\)

(b) Set the simplified fraction equal to 2:

\(\frac{x^2+x+4}{x^2-x} = 2\)

\(x^2+x+4 = 2(x^2-x)\)

\(x^2+x+4 = 2x^2-2x\)

Rearranging gives:

\(x^2-3x-4 = 0\)

Factorizing the quadratic:

\((x-4)(x+1) = 0\)

Thus, \(x = 4\) or \(x = -1\).

(a)
M1: Attempting to write over a common denominator, e.g., \(x(x+5) - 4(x-1)\).
M1: Expanding and simplifying the numerator to get \(x^2+x+4\).
A1: Correct final fraction of \(\frac{x^2+x+4}{x(x-1)}\).

(b)
M1: Equating their fraction to 2 and cross-multiplying.
M1: Forming a quadratic equation equal to 0, e.g., \(x^2 - 3x - 4 = 0\).
M1: Solving the quadratic equation by factoring or using the quadratic formula.
A1: \(x = 4\).
A1: \(x = -1\).

(a) Total volume is the sum of the volume of the cone and the volume of the hemisphere:

\(V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3\)

We are given that \(V = 240\pi\):

\(\frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = 240\pi\)

Divide through by \(\pi\):

\(\frac{1}{3}r^2 h + \frac{2}{3}r^3 = 240\)

Multiply both sides by 3:

\(r^2 h + 2r^3 = 720\)

\(r^2 h = 720 - 2r^3\)

\(h = \frac{720 - 2r^3}{r^2}\)

(b) Substitute \(r = 6\) into the formula for \(h\):

\(h = \frac{720 - 2(6^3)}{6^2} = \frac{720 - 432}{36} = \frac{288}{36} = 8\text{ cm}\).

The slant height \(l\) of the cone is given by:

\(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = 10\text{ cm}\).

The total surface area of the solid is the curved surface area of the cone plus the curved surface area of the hemisphere:

\(A = \pi r l + 2\pi r^2\)

\(A = \pi (6)(10) + 2\pi (6)^2 = 60\pi + 72\pi = 132\pi\text{ cm}^2\).

(a)
M1: Writing a correct equation for total volume: \(\frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = 240\pi\).
M1: Rearranging to make the term with \(h\) the subject.
A1: Correctly showing the algebraic steps to reach \(h = \frac{720 - 2r^3}{r^2}\).

(b)
M1: Substituting \(r = 6\) to find \(h = 8\).
M1: Using Pythagoras' Theorem to find the slant height \(l = 10\).
M1: Setting up the total surface area formula \(A = \pi r l + 2\pi r^2\) with their values.
A1: Obtaining the correct simplified answer of \(132\pi\).

(a) The total number of marbles is \(n + 4\).

The probability of drawing a red marble first is \(\frac{n}{n+4}\).

The probability of drawing a red marble second is \(\frac{n-1}{n+3}\).

The probability both are red is:

\(\text{P}(\text{Red, Red}) = \frac{n(n-1)}{(n+4)(n+3)}\)

(b) Given that this probability is \(\frac{1}{3}\):

\(\frac{n(n-1)}{(n+4)(n+3)} = \frac{1}{3}\)

\(3(n^2 - n) = (n+4)(n+3)\)

\(3n^2 - 3n = n^2 + 7n + 12\)

\(2n^2 - 10n - 12 = 0\)

(c) Divide the equation by 2:

\(n^2 - 5n - 6 = 0\)

\((n-6)(n+1) = 0\)

Since \(n\) must be positive, \(n = 6\).

With \(n = 6\), there are 6 red and 4 blue marbles (total of 10).

The probability of drawing different colours is:

\(\text{P}(\text{Red, Blue}) + \text{P}(\text{Blue, Red}) = \left(\frac{6}{10} \times \frac{4}{9}\right) + \left(\frac{4}{10} \times \frac{6}{9}\right)\)

\(= \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}\).

(a)
M1: Identifying \(n+4\) as the total number of marbles and stating a correct probability for one red draw.
A1: Correct product expression \(\frac{n(n-1)}{(n+4)(n+3)}\).

(b)
M1: Setting the product equal to \(\frac{1}{3}\) and attempting to cross-multiply.
M1: Correctly expanding both sides to get \(3n^2 - 3n\) and \(n^2 + 7n + 12\).
A1: Showing final algebraic simplification to \(2n^2 - 10n - 12 = 0\).

(c)
M1: Solving the quadratic equation to find \(n = 6\) (rejecting \(n = -1\)).
M1: Correctly setting up the probability sum for different colours with \(n = 6\).
A1: Correct final probability of \(\frac{8}{15}\) (or equivalent).

(a) The profit is modelled as a geometric sequence with initial term \(a = 20000\) and common ratio \(r = 1.08\).

Profit in Year 3 = \(20000 \times 1.08^2 = 23328\) (or £23,328).

(b) We require \(20000 \times 1.08^{n-1} > 40000\).

\(1.08^{n-1} > 2\)

Taking logarithms on both sides:

\((n-1) \ln(1.08) > \ln(2)\)

\(n-1 > \frac{\ln(2)}{\ln(1.08)}\)

\(n-1 > 9.006\)

\(n > 10.006\)

Since \(n\) must be an integer, the first year is Year 11.

(c) The total profits from Year 1 to Year 10 is the sum of the first 10 terms of the geometric series:

\(S_{10} = \frac{20000(1.08^{10} - 1)}{1.08 - 1}\)

\(S_{10} = \frac{20000(2.158925 - 1)}{0.08} = 250000 \times 1.158925 = 289731.25\).

Rounding to the nearest £100 gives £289,700.

(a)
M1: Calculating \(20000 \times 1.08^2\) or finding Year 2 profit (£21,600) and then multiplying by 1.08 again.
A1: £23,328.

(b)
M1: Setting up the inequality \(20000 \times 1.08^{n-1} > 40000\) (or using \(1.08^k > 2\)).
M1: Using logarithms or systematic trial and error to locate the boundary value.
A1: Year 11.

(c)
M1: Recalling and substituting the correct values into the geometric series sum formula with \(n = 10\).
M1: Evaluating the expression to reach an unrounded total of approx. 289731.
A1: Correctly rounding their answer to the nearest £100 (289,700).

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

Multiply both sides by the denominator:

\(y(2x-1) = x+2\)

\(2xy - y = x + 2\)

Group terms containing \(x\) on one side:

\(2xy - x = y + 2\)

\(x(2y - 1) = y + 2\)

\(x = \frac{y+2}{2y-1}\)

Therefore, \(\text{g}^{-1}(x) = \frac{x+2}{2x-1}\).

(b) First evaluate \(\text{f}(3)\):

\(\text{f}(3) = 3(3) - 5 = 4\).

Now evaluate \(\text{g}(4)\):

\(\text{g}(4) = \frac{4+2}{2(4)-1} = \frac{6}{7}\).

(c) Find \(\text{g}(2)\):

\(\text{g}(2) = \frac{2+2}{2(2)-1} = \frac{4}{3}\).

Set \(\text{f}(x) = \text{g}(2)\):

\(3x - 5 = \frac{4}{3}\)

\(3x = \frac{4}{3} + 5 = \frac{19}{3}\)

\(x = \frac{19}{9}\).

(a)
M1: Writing \(y = \frac{x+2}{2x-1}\) and attempting to multiply both sides by \(2x - 1\).
M1: Grouping \(x\) terms on one side and factorizing.
A1: Correctly stating \(\text{g}^{-1}(x) = \frac{x+2}{2x-1}\).

(b)
M1: Correctly substituting 3 into \(\text{f}(x)\) to get 4.
A1: Substituting their 4 into \(\text{g}(x)\) to get \(\frac{6}{7}\) (or equivalent decimal).

(c)
M1: Correctly evaluating \(\text{g}(2) = \frac{4}{3}\).
M1: Equating \(3x - 5 = \frac{4}{3}\) and attempting to solve for \(x\).
A1: Correct value of \(x = \frac{19}{9}\) (or \(2\frac{1}{9}\)).

(a) First find \(\overrightarrow{AB}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}\).

Since \(AQ : QB = 1 : 2\), \(\overrightarrow{AQ} = \frac{1}{3}\overrightarrow{AB} = \frac{1}{3}(\mathbf{b} - \mathbf{a})\).

Now express \(\overrightarrow{OQ}\):

\(\overrightarrow{OQ} = \overrightarrow{OA} + \overrightarrow{AQ} = \mathbf{a} + \frac{1}{3}(\mathbf{b} - \mathbf{a}) = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}\).

(b) Since \(OP : PA = 2 : 1\), \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).

\(\overrightarrow{BP} = \overrightarrow{BO} + \overrightarrow{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} - \mathbf{b}\).

(c) We can write \(\overrightarrow{OX}\) in two ways:

Method 1: Using line \(OQ\)

\(\overrightarrow{OX} = k \overrightarrow{OQ} = k\left(\frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}\right) = \frac{2k}{3}\mathbf{a} + \frac{k}{3}\mathbf{b}\)

Method 2: Using line \(BP\)

\(\overrightarrow{OX} = \overrightarrow{OB} + \overrightarrow{BX} = \mathbf{b} + m \overrightarrow{BP}\)

\(\overrightarrow{OX} = \mathbf{b} + m\left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2m}{3}\mathbf{a} + (1-m)\mathbf{b}\)

Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):

For \(\mathbf{a}\): \(\frac{2k}{3} = \frac{2m}{3} \implies k = m\)

For \(\mathbf{b}\): \(\frac{k}{3} = 1 - m\)

Substitute \(m = k\) into the second equation:

\(\frac{k}{3} = 1 - k \implies \frac{4k}{3} = 1 \implies k = \frac{3}{4}\)

Thus, \(k = \frac{3}{4}\) and \(m = \frac{3}{4}\).

(a)
M1: Finding a correct expression for \(\overrightarrow{AB}\) or \(\overrightarrow{AQ}\).
A1: Correctly simplifying to obtain \(\overrightarrow{OQ} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}\).

(b)
M1: Identifying \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).
A1: Correctly stating \(\overrightarrow{BP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\).

(c)
M1: Expressing \(\overrightarrow{OX}\) in terms of \(k\) and their \(\overrightarrow{OQ}\).
M1: Expressing \(\overrightarrow{OX}\) in terms of \(m\) and their \(\overrightarrow{BP}\).
M1: Comparing coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to form simultaneous equations.
A1: Correctly finding both \(k = \frac{3}{4}\) and \(m = \frac{3}{4}\).

(a) Let the North-line at \(P\) and the North-line at \(Q\) be parallel.

The bearing of \(Q\) from \(P\) is \(060^\circ\).

By co-interior angles, the angle of the line \(PQ\) with the North-pointing line at \(Q\) is \(180^\circ - 60^\circ = 120^\circ\) (measured counterclockwise).

The bearing of \(R\) from \(Q\) is \(150^\circ\) (measured clockwise from North).

Therefore, the angle \(\angle PQR = 360^\circ - (120^\circ + 150^\circ) = 90^\circ\).

Since \(\angle PQR = 90^\circ\), triangle \(PQR\) is a right-angled triangle. We can use Pythagoras' Theorem to find \(PR\):

\(PR = \sqrt{PQ^2 + QR^2} = \sqrt{15^2 + 25^2} = \sqrt{225 + 625} = \sqrt{850} \approx 29.155\text{ km}\).

To 3 significant figures, \(PR = 29.2\text{ km}\).

(b) Since the triangle is right-angled at \(Q\), the angle \(\angle QPR\) can be found using trigonometry:

\(\tan(\angle QPR) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{QR}{PQ} = \frac{25}{15} = \frac{5}{3}\)

\(\angle QPR = \arctan\left(\frac{5}{3}\right) \approx 59.04^\circ\).

The bearing of \(R\) from \(P\) is the sum of the bearing of \(Q\) from \(P\) and the angle \(\angle QPR\):

Bearing of \(R\) from \(P\) = \(60^\circ + 59.04^\circ = 119.04^\circ\).

To the nearest degree, this is \(119^\circ\).

(a)
M1: Attempting to find the angle \(\angle PQR\) using parallel lines/bearings.
A1: Finding that \(\angle PQR = 90^\circ\) (or using the Cosine Rule with this angle).
M1: Using Pythagoras' Theorem: \(PR = \sqrt{15^2 + 25^2}\).
A1: Correctly rounding \(PR\) to 29.2 (allow unrounded answers from 29.15 to 29.2).

(b)
M1: Attempting to use a trigonometric ratio to find angle \(\angle QPR\), e.g., \(\tan(\angle QPR) = \frac{25}{15}\).
A1: Finding \(\angle QPR \approx 59.0^\circ\) (or 59.04).
M1: Adding their angle to the bearing of \(60^\circ\).
A1: Correct bearing of \(119^\circ\) (accept 119).

(a) We first calculate the number of students who play exactly two sports:
- Football and Basketball only: \(12 - 5 = 7\).
- Basketball and Tennis only: \(9 - 5 = 4\).
- Football and Tennis only: \(10 - 5 = 5\).
The other regions are given as:
- Only Football: \(2x\)
- Only Basketball: \(x+3\)
- Only Tennis: \(x\)
- None: \(8\)
- Intersection of all three: \(5\)
Draw three intersecting circles inside a rectangular universal set, and place these numbers/expressions in their respective regions.

(b) The sum of the number of students in all 8 regions must equal the total number of students (80):
\(2x + (x+3) + x + 7 + 4 + 5 + 5 + 8 = 80\)

Simplify the equation:
\(4x + 32 = 80\)

\(4x = 48 \implies x = 12\).

(c) The students who play at least two sports are those in the regions: \(F \cap B \text{ only}\), \(B \cap T \text{ only}\), \(F \cap T \text{ only}\), and \(F \cap B \cap T\).

Number of such students = \(7 + 4 + 5 + 5 = 21\).

Since there are 80 students in total, the probability is \(\frac{21}{80}\) (or 0.2625).

(a)
M1: Drawing three intersecting circles with a central region of 5.
M1: Calculating the intersection-only values of 7, 4, and 5.
A1: Completely correct Venn diagram with all 8 regions correctly labeled.

(b)
M1: Summing up all 8 regions and equating the expression to 80.
A1: Correctly simplifying the expression to show \(4x + 32 = 80\) and obtaining \(x = 12\).

(c)
M1: Identifying the correct regions for 'at least two sports' (7, 4, 5, 5).
M1: Adding these values together to get 21.
A1: Correct probability of \(\frac{21}{80}\) (or equivalent decimal 0.2625).

(a) The time taken for the outward journey is given by \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\):
\text{Time} = \frac{40}{x} \text{ hours}

(b) On the return journey, the speed is \(x + 3\) km/h. The time taken is:
\text{Time} = \frac{40}{x+3} \text{ hours}

(c) The difference in time is 40 minutes, which is \(\frac{40}{60} = \frac{2}{3}\) hours:
\frac{40}{x} - \frac{40}{x+3} = \frac{2}{3}

Divide the entire equation by 2:
\frac{20}{x} - \frac{20}{x+3} = \frac{1}{3}

Multiply by the common denominator \(3x(x+3)\):
20(3)(x+3) - 20(3)x = x(x+3)
60x + 180 - 60x = x^2 + 3x
180 = x^2 + 3x
x^2 + 3x - 180 = 0

(d) Solve the quadratic equation by factoring:
(x - 12)(x + 15) = 0
So, \(x = 12\) or \(x = -15\).
Since speed must be positive, \(x = 12\) km/h.

(a) B1: For the correct expression \(\frac{40}{x}\) (1 mark)

(b) B1: For the correct expression \(\frac{40}{x+3}\) (1 mark)

(c) M1: For setting up the correct equation with their expressions: \(\frac{40}{x} - \frac{40}{x+3} = \frac{2}{3}\) (or 40 minutes equated appropriately in hours)
M1: For algebraic steps to eliminate fractions, e.g., \(60(x+3) - 60x = x(x+3)\)
A1: Correctly simplifying to the given quadratic equation \(x^2 + 3x - 180 = 0\) with no errors shown (3 marks total for c)

(d) M1: For attempting to factorise or solve using the quadratic formula: \((x-12)(x+15) = 0\) (or equivalent)
A1: Finding the roots \(x = 12\) and \(x = -15\)
A1ft: Rejecting the negative root and stating \(12\) km/h as the average outward speed (3.33 marks total for d)

(a) The probability of choosing a red bead first is \(\frac{5}{n}\).
Since the first bead is not replaced, there are now \(n-1\) beads left, and 4 of them are red.
The probability of choosing a second red bead is \(\frac{4}{n-1}\).
The probability that both are red is:
\text{P(Red, Red)} = \frac{5}{n} \times \frac{4}{n-1} = \frac{20}{n(n-1)}
We are given that this probability is \(\frac{5}{33}\):
\frac{20}{n(n-1)} = \frac{5}{33}
Multiply both sides by \(n(n-1)\) and 33:
20 \times 33 = 5n(n-1)
660 = 5(n^2 - n)
Divide by 5:
132 = n^2 - n
n^2 - n - 132 = 0

(b) Solve the quadratic equation by factoring:
(n - 12)(n + 11) = 0
So, \(n = 12\) or \(n = -11\).
Since the total number of beads must be positive, \(n = 12\).

(c) When \(n = 12\), the total number of beads is 12, with 5 red beads and \(12 - 5 = 7\) blue beads.
The probability of choosing beads of different colours is:
\text{P(Red, Blue)} + \text{P(Blue, Red)}
= \left(\frac{5}{12} \times \frac{7}{11}\right) + \left(\frac{7}{12} \times \frac{5}{11}\right)
= \frac{35}{132} + \frac{35}{132}
= \frac{70}{132} = \frac{35}{66}

(a) M1: For writing a correct expression for the probability of both red, e.g., \(\frac{5}{n} \times \frac{4}{n-1}\)
M1: Equating their expression to \(\frac{5}{33}\) and showing algebraic manipulation to remove denominators, e.g., \(20 \times 33 = 5n(n-1)\)
A1: Showing clear intermediate steps to obtain the given quadratic equation: \(n^2 - n - 132 = 0\) (3 marks total for a)

(b) M1: Attempting to factorise or solve \(n^2 - n - 132 = 0\), e.g., \((n-12)(n+11) = 0\)
A1: Rejecting \(n = -11\) and stating \(n = 12\) (2 marks total for b)

(c) M1: Finding the number of blue beads as \(12 - 5 = 7\), and setting up the sum of probabilities for different colours: \(\text{P(R, B)} + \text{P(B, R)}\)
M1: For evaluating the calculation: \(\left(\frac{5}{12} \times \frac{7}{11}\right) + \left(\frac{7}{12} \times \frac{5}{11}\right)\)
A1: Providing the correct final fraction \(\frac{35}{66}\) (or equivalent decimal \(0.530\) to 3 s.f.) (3.33 marks total for c)

(a) The total surface area \(A\) of a solid cylinder is:
A = 2\pi r^2 + 2\pi r h
Given \(A = 150\pi\):
2\pi r^2 + 2\pi r h = 150\pi
Divide every term by \(2\pi\):
r^2 + r h = 75
r h = 75 - r^2
h = \frac{75 - r^2}{r}

(b) The volume \(V\) of a cylinder is given by \(V = \pi r^2 h\).
Substituting the expression for \(h\) from part (a):
V = \pi r^2 \left(\frac{75 - r^2}{r}\right)
V = \pi r (75 - r^2)
V = 75\pi r - \pi r^3

(c) To find the maximum volume, differentiate \(V\) with respect to \(r\):
\frac{dV}{dr} = 75\pi - 3\pi r^2
Set the derivative equal to zero for a stationary point:
75\pi - 3\pi r^2 = 0
3\pi r^2 = 75\pi
r^2 = 25
Since the radius must be positive, \(r = 5\) cm.

Now calculate the maximum volume by substituting \(r = 5\) into the volume formula:
V = 75\pi(5) - \pi(5)^3 = 375\pi - 125\pi = 250\pi
V \approx 785.398... \text{ cm}^3
To 3 significant figures, the maximum volume is 785 \(\text{cm}^3\).

(a) M1: Recalling the correct total surface area formula \(2\pi r^2 + 2\pi r h = 150\pi\) (or equivalent)
M1: Rearranging the formula to isolate the term with \(h\), e.g., \(2\pi r h = 150\pi - 2\pi r^2\)
A1: Showing clear algebraic steps to arrive at \(h = \frac{75 - r^2}{r}\) with no errors (3 marks total for a)

(b) M1: Recalling the correct volume formula \(V = \pi r^2 h\) and substituting their expression for \(h\)
A1: Showing simplification to obtain the given expression \(V = 75\pi r - \pi r^3\) (2 marks total for b)

(c) M1: Correctly differentiating \(V\) to find \(\frac{dV}{dr} = 75\pi - 3\pi r^2\)
M1: Setting \(\frac{dV}{dr} = 0\) and solving for \(r\) (yielding \(r = 5\))
A1: Obtaining \(r = 5\) cm (or shown with second derivative check, though not strictly required here)
A1: Calculating the maximum volume as \(250\pi\) and rounding to 3 s.f. to get \(785\) \(\text{cm}^3\) (3.33 marks total for c)

(a)
(i) Since \(OP : PA = 2 : 1\), the point \(P\) is \(\frac{2}{3}\) of the way along \(OA\). Therefore, \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).
\\ \overrightarrow{BP} = \overrightarrow{BO} + \overrightarrow{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} - \mathbf{b}

(ii) Since \(Q\) is the midpoint of \(AB\):
\\ \overrightarrow{OQ} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}

(b)
First expression for \(\overrightarrow{OX}\):
\\ \overrightarrow{OX} = \mu \overrightarrow{OQ} = \frac{1}{2}\mu \mathbf{a} + \frac{1}{2}\mu \mathbf{b}

Second expression for \(\overrightarrow{OX}\):
\\ \overrightarrow{OX} = \overrightarrow{OB} + \overrightarrow{BX} = \mathbf{b} + \lambda \overrightarrow{BP} = \mathbf{b} + \lambda \left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\lambda \mathbf{a} + (1 - \lambda)\mathbf{b}

(c) Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) from the two expressions in part (b):
For \(\mathbf{a}\):
\frac{1}{2}\mu = \frac{2}{3}\lambda \implies \mu = \frac{4}{3}\lambda

For \(\mathbf{b}\):
\frac{1}{2}\mu = 1 - \lambda

Substitute \(\mu = \frac{4}{3}\lambda\) into the second equation:
\frac{1}{2}\left(\frac{4}{3}\lambda\right) = 1 - \lambda
\frac{2}{3}\lambda = 1 - \lambda
\frac{5}{3}\lambda = 1 \implies \lambda = \frac{3}{5}

Now find \(\mu\):
\mu = \frac{4}{3}\left(\frac{3}{5} ight) = \frac{4}{5}

(a) B1: For \(\overrightarrow{BP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\) (or equivalent)
B1: For \(\overrightarrow{OQ} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\) (or equivalent) (2 marks total for a)

(b) B1: For writing \(\overrightarrow{OX} = \frac{1}{2}\mu \mathbf{a} + \frac{1}{2}\mu \mathbf{b}\)
M1: For starting the second path \(\overrightarrow{OX} = \overrightarrow{OB} + \lambda \overrightarrow{BP}\)
A1: Correctly simplifying to \(\overrightarrow{OX} = \frac{2}{3}\lambda \mathbf{a} + (1 - \lambda)\mathbf{b}\) (3 marks total for b)

(c) M1: Setting up simultaneous equations by equating coefficients, e.g., \(\frac{1}{2}\mu = \frac{2}{3}\lambda\) and \(\frac{1}{2}\mu = 1 - \lambda\)
M1: Attempting to solve the simultaneous equations for one variable, e.g., substituting \(\mu = \frac{4}{3}\lambda\) into the second equation
A1: Finding \(\lambda = \frac{3}{5}\) (or equivalent decimal \(0.6\))
A1ft: Finding \(\mu = \frac{4}{5}\) (or equivalent decimal \(0.8\)) (3.33 marks total for c)