2023 Edexcel IGCSE Mathematics (Specification B) Practice Paper with Answers

Since \(\text{n}(A \cap B)' = 33\) and \(\text{n}(\mathscr{U}) = 40\), we have \(\text{n}(A \cap B) = \text{n}(\mathscr{U}) - \text{n}(A \cap B)' = 40 - 33 = 7\). Now, using the formula for the union of two sets, \(\text{n}(A \cup B) = \text{n}(A) + \text{n}(B) - \text{n}(A \cap B)\), we get \(\text{n}(A \cup B) = 22 + 15 - 7 = 30\). Finally, the number of elements in the complement of the union is \(\text{n}(A \cup B)' = \text{n}(\mathscr{U}) - \text{n}(A \cup B) = 40 - 30 = 10\).

M1: For finding \(\text{n}(A \cap B) = 7\) or writing \(\text{n}(A \cup B) = 22 + 15 - \text{n}(A \cap B)\) with correct values. A1: Correct final answer of 10.

First, calculate the original selling price before the sale: Original selling price = \(80 \times 1.15 = £92\). Next, calculate the selling price during the sale after a 10% reduction: Sale price = \(92 \times (1 - 0.10) = 92 \times 0.9 = £82.80\). Now, find the profit made during the sale: Profit = \(82.80 - 80 = £2.80\). Finally, calculate the percentage profit on the cost price: Percentage profit = \(\frac{2.80}{80} \times 100\% = 3.5\%\).

M1: For calculating the original selling price as £92 or writing a single expression for the sale price \(80 \times 1.15 \times 0.9\). M1: For calculating the sale price as £82.80. A1: For the correct percentage profit of 3.5%.

The determinant of a \(2 \times 2\) matrix is given by \(ad - bc\). Therefore, the determinant of \(\mathbf{M}\) is \(\det(\mathbf{M}) = x(x - 1) - (3)(4) = x^2 - x - 12\). We are given that \(\det(\mathbf{M}) = 18\), so \(x^2 - x - 12 = 18\) which simplifies to \(x^2 - x - 30 = 0\). Factorising the quadratic equation gives \((x - 6)(x + 5) = 0\). This gives \(x = 6\) or \(x = -5\). Since \(x\) must be a positive value, we select \(x = 6\).

M1: For expressing the determinant as \(x(x - 1) - 12\) or \(x^2 - x - 12\). M1: For setting up the quadratic equation \(x^2 - x - 30 = 0\) and attempting to solve it. A1: For \(x = 6\) (rejecting \(x = -5\)).

To express the subtraction as a single fraction, find a common denominator, which is \((2x - 1)(3x + 2)\). This gives \(\frac{3(3x + 2) - 2(2x - 1)}{(2x - 1)(3x + 2)}\). Expand the terms in the numerator: \(\frac{9x + 6 - 4x + 2}{(2x - 1)(3x + 2)}\). Simplify the numerator by combining like terms to get \(\frac{5x + 8}{(2x - 1)(3x + 2)}\).

M1: For putting the fractions over a common denominator with at least one correct numerator term expanded or indicated: \(3(3x + 2) - 2(2x - 1)\). M1: For expanding the numerator correctly to \(9x + 6 - 4x + 2\). A1: For the correct final simplified fraction \(\frac{5x+8}{(2x-1)(3x+2)}\) (accept with expanded denominator \(6x^2 + x - 2\)).

First, find the vector \(\vec{AB} = \mathbf{b} - \mathbf{a} = (8\mathbf{i} + 8\mathbf{j}) - (2\mathbf{i} - \mathbf{j}) = 6\mathbf{i} + 9\mathbf{j}\). Since \(AP:PB = 1:2\), the point \(P\) is one-third of the way along \(AB\) from \(A\), so \(\vec{AP} = \frac{1}{3}\vec{AB} = \frac{1}{3}(6\mathbf{i} + 9\mathbf{j}) = 2\mathbf{i} + 3\mathbf{j}\). Now, find the position vector of \(P\): \(\vec{OP} = \vec{OA} + \vec{AP} = (2\mathbf{i} - \mathbf{j}) + (2\mathbf{i} + 3\mathbf{j}) = 4\mathbf{i} + 2\mathbf{j}\).

M1: For finding \(\vec{AB} = 6\mathbf{i} + 9\mathbf{j}\) or setting up the vector equation \(\vec{OP} = \mathbf{a} + \frac{1}{3}(\mathbf{b} - \mathbf{a})\). A1: For the correct vector \(4\mathbf{i} + 2\mathbf{j}\).

First, evaluate the inner function \(\text{f}(x)\) at \(x = 5\): \(\text{f}(5) = 2(5) - 3 = 10 - 3 = 7\). Now, substitute this result into \(\text{g}(x)\): \(\text{gf}(5) = \text{g}(7) = 7^2 + 1 = 49 + 1 = 50\).

M1: For correctly evaluating \(\text{f}(5) = 7\) or for finding the algebraic expression for \(\text{gf}(x) = (2x - 3)^2 + 1\). A1: For the correct final value of 50.

The total volume is the sum of the volume of the cylinder and the volume of the hemisphere. Volume of the cylinder is \(V_{\text{cylinder}} = \pi r^2 h = \pi \times 3^2 \times 8 = 72\pi\text{ cm}^3\). Volume of the hemisphere is \(V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi \times 3^3 = 18\pi\text{ cm}^3\). Total volume is \(V_{\text{total}} = 72\pi + 18\pi = 90\pi\text{ cm}^3\).

M1: For calculating either the volume of the cylinder (\(72\pi\)) or the volume of the hemisphere (\(18\pi\)) correctly in terms of \(\pi\). A1: For the correct total volume of \(90\pi\).

The area of a triangle can be found using the formula \(\text{Area} = \frac{1}{2} b c \sin A\). Substitute the given values into the formula: \(17.5 = \frac{1}{2} \times 10 \times 7 \times \sin(\angle BAC)\), which simplifies to \(17.5 = 35 \sin(\angle BAC)\). Solving for \(\sin(\angle BAC)\) gives \(\sin(\angle BAC) = \frac{17.5}{35} = 0.5\). The acute angle with a sine of \(0.5\) is \(30^\circ\). Since angle \(BAC\) is obtuse (\(90^\circ < \angle BAC < 180^\circ\)), we have \(\angle BAC = 180^\circ - 30^\circ = 150^\circ\).

M1: For using the area formula to write \(17.5 = \frac{1}{2} \times 10 \times 7 \times \sin(\angle BAC)\). M1: For finding \(\sin(\angle BAC) = 0.5\) and identifying either \(30^\circ\) or the obtuse angle calculation. A1: For the correct angle of \(150^\circ\).

First, list the elements of the universal set and the sets \(A\) and \(B\):
\(\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\)
\(A = \{3, 6, 9, 12\}\)
\(B = \{1, 2, 3, 4, 6, 12\}\)

The complement of \(B\), denoted as \(B'\), consists of elements in \(\mathcal{U}\) that are not in \(B\):
\(B' = \{5, 7, 8, 9, 10, 11\}\)

Now, find the intersection \(A \cap B'\), which consists of elements that are in both \(A\) and \(B'\):
\(A \cap B' = \{9\}\)

The number of elements in this intersection is \(n(A \cap B') = 1\).

M1 for correctly listing the elements of set \(A\) and set \(B\) (or \(B'\)).
A1 for 1 (correct final answer).

Let \(x = 0.2\dot{3}\dot{5} = 0.2353535...\)

Multiply by 10 to move the non-repeating part past the decimal point:
\(10x = 2.353535...\) [Equation 1]

Multiply by 1000 to move one full repeating block past the decimal point:
\(1000x = 235.353535...\) [Equation 2]

Subtract Equation 1 from Equation 2:
\(1000x - 10x = 235.353535... - 2.353535...\)
\(990x = 233\)

Solve for \(x\):
\(x = \frac{233}{990}\)

Since 233 is a prime number and does not divide 990, the fraction is in its simplest form.

M1 for setting up two equations that can be subtracted to eliminate the recurring decimal (e.g., finding \(10x\) and \(1000x\)).
M1 for performing the subtraction to obtain \(990x = 233\) (or equivalent).
A1 for \(\frac{233}{990}\).

First, compute the matrix product \(\mathbf{A}\mathbf{B}\):
\(\mathbf{A}\mathbf{B} = \begin{pmatrix} 2 & p \\ -1 & 3 \end{pmatrix} \begin{pmatrix} 4 \\ 1 \end{pmatrix} = \begin{pmatrix} 2(4) + p(1) \\ -1(4) + 3(1) \end{pmatrix} = \begin{pmatrix} 8 + p \\ -1 \end{pmatrix}\)

We are given that \(\mathbf{A}\mathbf{B} = \begin{pmatrix} 13 \\ q \end{pmatrix}\). Equating the corresponding elements:
1) \(8 + p = 13 \implies p = 5\)
2) \(q = -1\)

M1 for attempting the matrix multiplication to get at least one correct element expression: \(8+p\) or \(-1\).
A1 for \(p = 5\).
A1 for \(q = -1\).

To subtract the fractions, find a common denominator, which is \((x-2)(x+1)\):
\(\frac{3}{x-2} - \frac{2}{x+1} = \frac{3(x+1) - 2(x-2)}{(x-2)(x+1)}

Now expand and simplify the numerator:
\)3(x+1) - 2(x-2) = 3x + 3 - 2x + 4 = x + 7\)

So the single simplified fraction is:
\(\frac{x+7}{(x-2)(x+1)}\)

M1 for writing both terms with a common denominator, i.e. \(\frac{3(x+1) - 2(x-2)}{(x-2)(x+1)}\) (allow 1 sign error in the expansion of the numerator).
A1 for \(\frac{x+7}{(x-2)(x+1)}\) or \(\frac{x+7}{x^2-x-2}\).

Use the Cosine Rule to find the length of \(BC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC)\)

Substitute the given values into the formula:
\(BC^2 = 5^2 + 8^2 - 2(5)(8)\cos(60^\circ)\)
\(BC^2 = 25 + 64 - 80(0.5)\)
\(BC^2 = 89 - 40\)
\(BC^2 = 49\)

Take the square root of both sides:
\(BC = \sqrt{49} = 7\text{ cm}\)

M1 for substituting the correct values into the Cosine Rule: \(5^2 + 8^2 - 2(5)(8)\cos(60^\circ)\).
M1 for simplifying the expression to find \(BC^2 = 49\).
A1 for \(7\) (or \(7\text{ cm}\)).

Multiply the entire equation by \(x(x-1)\) to clear the fractions: \(6x - 4(x-1) = x(x-1)\). Expand both sides: \(6x - 4x + 4 = x^2 - x\), which simplifies to \(2x + 4 = x^2 - x\). Rearrange into a quadratic equation: \(x^2 - 3x - 4 = 0\). Factorise the quadratic: \((x-4)(x+1) = 0\). This gives the solutions: \(x = 4\) or \(x = -1\).

M1 for multiplying by \(x(x-1)\) to get \(6x - 4(x-1) = x(x-1)\) or equivalent. A1 for correct expansion and simplification to a quadratic equation, e.g., \(x^2 - 3x - 4 = 0\). M1 for attempting to factorise or solve their 3-term quadratic. A1 for one correct solution. A1 for both correct solutions.

First, calculate the product \(\mathbf{AB}\): \(\mathbf{AB} = \begin{pmatrix} a & 3 \\ -2 & b \end{pmatrix} \begin{pmatrix} 2 & -1 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} 2a + 12 & -a + 15 \\ -4 + 4b & 2 + 5b \end{pmatrix}\). Set this equal to the given product matrix: \(\begin{pmatrix} 2a + 12 & -a + 15 \\ -4 + 4b & 2 + 5b \end{pmatrix} = \begin{pmatrix} 18 & 12 \\ 12 & 22 \end{pmatrix}\). From the top-left entry, we have \(2a + 12 = 18 \Rightarrow 2a = 6 \Rightarrow a = 3\). From the bottom-left entry, we have \(-4 + 4b = 12 \Rightarrow 4b = 16 \Rightarrow b = 4\). Thus, \(a = 3\) and \(b = 4\).

M1 for attempting matrix multiplication to find at least two terms of \(\mathbf{AB}\) in terms of \(a\) and \(b\). A1 for a correct product matrix. M1 for setting up a linear equation for \(a\) or \(b\), e.g., \(2a + 12 = 18\). A1 for \(a = 3\). A1 for \(b = 4\).

(a) First find \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -3\mathbf{a} + 3\mathbf{b}\). Since \(P\) divides \(AB\) in the ratio \(2:1\), \(\overrightarrow{AP} = \frac{2}{3}\overrightarrow{AB} = \frac{2}{3}(-3\mathbf{a} + 3\mathbf{b}) = -2\mathbf{a} + 2\mathbf{b}\). Therefore, \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = 3\mathbf{a} + (-2\mathbf{a} + 2\mathbf{b}) = \mathbf{a} + 2\mathbf{b}\). (b) Since \(Q\) divides \(OB\) in the ratio \(1:2\), \(\overrightarrow{OQ} = \frac{1}{3}\overrightarrow{OB} = \mathbf{b}\). Then, \(\overrightarrow{QP} = \overrightarrow{QO} + \overrightarrow{OP} = -\mathbf{b} + (\mathbf{a} + 2\mathbf{b}) = \mathbf{a} + \mathbf{b}\).

M1 for finding \(\overrightarrow{AB} = -3\mathbf{a} + 3\mathbf{b}\). A1 for \(\overrightarrow{OP} = \mathbf{a} + 2\mathbf{b}\). M1 for identifying \(\overrightarrow{OQ} = \mathbf{b}\). M1 for writing a correct vector path for \(\overrightarrow{QP}\), e.g., \(\overrightarrow{QO} + \overrightarrow{OP}\). A1 for \(\overrightarrow{QP} = \mathbf{a} + \mathbf{b}\).

(a) Let \(y = \frac{2x+5}{x-1}\). Rearrange to make \(x\) the subject: \(y(x-1) = 2x+5 \Rightarrow xy - y = 2x+5 \Rightarrow xy - 2x = y+5 \Rightarrow x(y-2) = y+5 \Rightarrow x = \frac{y+5}{y-2}\). Thus, \(f^{-1}(x) = \frac{x+5}{x-2}\). (b) If \(fg(x) = 3\), then \(g(x) = f^{-1}(3)\). Using the expression from part (a): \(f^{-1}(3) = \frac{3+5}{3-2} = 8\). Since \(g(x) = 2x-4\), we have \(2x-4 = 8 \Rightarrow 2x = 12 \Rightarrow x = 6\).

M1 for setting \(y = \frac{2x+5}{x-1}\) and attempting to clear the fraction. M1 for successfully grouping terms in \(x\) on one side and factorising, e.g., \(x(y-2) = y+5\). A1 for \(f^{-1}(x) = \frac{x+5}{x-2}\). M1 for equating \(2x-4\) to \(f^{-1}(3)\) or expanding \(fg(x)\) and attempting to solve. A1 for \(x = 6\).

Let the multiplier for compound interest be \(x = 1 + \frac{r}{100}\). After 2 years, the investment value is \(P x^2 = 8820\). After 3 years, the investment value is \(P x^3 = 9261\). Divide the second equation by the first equation: \(\frac{P x^3}{P x^2} = \frac{9261}{8820} \Rightarrow x = 1.05\). Since \(1 + \frac{r}{100} = 1.05\), we get \(r = 5\). Substitute \(x = 1.05\) back into the first equation: \(P(1.05)^2 = 8820 \Rightarrow P(1.1025) = 8820 \Rightarrow P = \frac{8820}{1.1025} = 8000\). Thus, \(r = 5\) and \(P = 8000\).

M1 for writing down two equations representing compound interest. M1 for dividing the two equations to eliminate \(P\) and find the multiplier. A1 for \(r = 5\). M1 for substituting their value of \(r\) back into one of the equations to find \(P\). A1 for \(P = 8000\).

(a) The probability of drawing the first red marble is \(\frac{6}{n}\). Since the marble is not replaced, the probability of drawing the second red marble is \(\frac{5}{n-1}\). The probability that both are red is \(\frac{6}{n} \times \frac{5}{n-1} = \frac{30}{n(n-1)}\triangle\). We are given this probability is \(\frac{1}{3}\), so \(\frac{30}{n(n-1)} = \frac{1}{3} \Rightarrow n(n-1) = 90 \Rightarrow n^2 - n - 90 = 0\). (b) Solve the quadratic equation: \((n - 10)(n + 9) = 0\). Since \(n\) must be positive, \(n = 10\). The number of blue marbles is the total number of marbles minus the number of red marbles: \(10 - 6 = 4\).

M1 for expressing the probability of selecting two red marbles without replacement as \(\frac{6}{n} \times \frac{5}{n-1}\). M1 for setting this expression equal to \(\frac{1}{3}\). A1 for completing the algebra to show \(n^2 - n - 90 = 0\) with no errors. M1 for factorising the quadratic to solve for \(n\), obtaining \(n = 10\). A1 for finding the correct number of blue marbles, which is 4.

The area of a triangle is given by \(\frac{1}{2} b c \sin A\). Here, \(\text{Area} = \frac{1}{2} \times 8 \times 11 \times \sin(BAC) = 44 \sin(BAC)\). We are given that the area is \(22\sqrt{3}\), so \(44 \sin(BAC) = 22\sqrt{3} \Rightarrow \sin(BAC) = \frac{\sqrt{3}}{2}\). The two possible values for angle \(BAC\) are \(60^\circ\) or \(120^\circ\). Since angle \(BAC\) is obtuse, \(BAC = 120^\circ\). Now use the Cosine Rule to find \(BC\): \(BC^2 = AB^2 + AC^2 - 2(AB)(AC) \cos(BAC)\). Substitute the values: \(BC^2 = 8^2 + 11^2 - 2(8)(11) \cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), we have \(BC^2 = 64 + 121 - 176(-0.5) = 185 + 88 = 273\). Therefore, \(BC = \sqrt{273} \approx 16.5227\text{ cm}\). To 3 significant figures, \(BC = 16.5\text{ cm}\).

M1 for setting up the area equation: \(\frac{1}{2} \times 8 \times 11 \times \sin(BAC) = 22\sqrt{3}\). A1 for finding \(\sin(BAC) = \frac{\sqrt{3}}{2}\) and identifying the angle as \(120^\circ\). M1 for stating the Cosine Rule: \(BC^2 = 8^2 + 11^2 - 2(8)(11)\cos(120^\circ)\). A1 for calculating \(BC^2 = 273\). A1 for \(BC = 16.5\) (or better, 16.52...).

(a) The volume of the solid is the sum of the volume of the cylinder and the volume of the hemisphere: \(V = \pi r^2 h + \frac{2}{3}\pi r^3\). Substitute \(V = 504\pi\) and \(h = 10\): \(504\pi = 10\pi r^2 + \frac{2}{3}\pi r^3\). Divide both sides by \(\pi\): \(504 = 10r^2 + \frac{2}{3}r^3\). Multiply the entire equation by 3 to clear the fraction: \(1512 = 30r^2 + 2r^3\), which simplifies to \(2r^3 + 30r^2 - 1512 = 0\). Divide by 2: \(r^3 + 15r^2 - 756 = 0\). (b) To show \(r = 6\) is a solution, substitute \(r = 6\) into the equation: \(6^3 + 15(6^2) - 756 = 216 + 540 - 756 = 0\). Hence, \(r = 6\) is a solution. The total surface area of the solid consists of the curved surface area of the hemisphere, the curved surface area of the cylinder, and the base area of the cylinder: \(A = 2\pi r^2 + 2\pi r h + \pi r^2 = 3\pi r^2 + 2\pi r h\). Substitute \(r = 6\) and \(h = 10\): \(A = 3\pi(6^2) + 2\pi(6)(10) = 108\pi + 120\pi = 228\pi\text{ cm}^2\).

M1 for writing down the correct volume formula: \(V = \pi r^2 h + \frac{2}{3}\pi r^3\) and substituting \(V = 504\pi\) and \(h = 10\). M1 for clearing the fraction and dividing by \(\pi\) to obtain \(2r^3 + 30r^2 - 1512 = 0\) or equivalent. A1 for completing the algebra to get \(r^3 + 15r^2 - 756 = 0\). B1 for verifying that \(6^3 + 15(6^2) - 756 = 0\). M1 for calculating total surface area as \(3\pi r^2 + 2\pi r h\). A1 for \(228\pi\).

Multiply both sides of the equation by the common denominator \((x - 3)(x - 1)\): \(9(x - 1) - 10(x - 3) = 2(x - 3)(x - 1)\). Expand the brackets: \(9x - 9 - 10x + 30 = 2(x^2 - 4x + 3)\). Simplify the terms: \(-x + 21 = 2x^2 - 8x + 6\). Rearrange the terms to form a quadratic equation: \(2x^2 - 7x - 15 = 0\). Factorise the quadratic expression: \((2x + 3)(x - 5) = 0\). Thus, the solutions are \(x = 5\) and \(x = -1.5\).

M1: For removing fractions by multiplying through by the common denominator, e.g., \(9(x-1) - 10(x-3) = 2(x-3)(x-1)\).
M1: For expanding brackets correctly, e.g., \(9x - 9 - 10x + 30 = 2x^2 - 8x + 6\).
A1: For obtaining a correct 3-term quadratic equation in the form \(2x^2 - 7x - 15 = 0\) (or any equivalent form).
M1: For attempting to solve their 3-term quadratic equation by factorisation, formula, or completing the square.
A1: For both correct solutions \(x = 5\) and \(x = -1.5\) (or \(x = -\frac{3}{2}\)).

Express the vectors \(\overrightarrow{OP}\) and \(\overrightarrow{OQ}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). Since \(P\) lies on \(OA\) and \(OP : PA = 2 : 1\), we have \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\). Since \(Q\) lies on \(AB\) and \(AQ : QB = 3 : 2\), we have \(\overrightarrow{OQ} = \overrightarrow{OA} + \frac{3}{5}\overrightarrow{AB} = \mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a}) = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\). Since \(X\) lies on the line \(OQ\), \(\overrightarrow{OX} = \mu \overrightarrow{OQ} = \mu \left(\frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\right)\). Since \(X\) lies on the line \(BP\), \(\overrightarrow{OX} = \overrightarrow{OB} + \lambda \overrightarrow{BP} = \mathbf{b} + \lambda \left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\lambda \mathbf{a} + (1 - \lambda)\mathbf{b}\). Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\): For \(\mathbf{a}\): \(\frac{2}{5}\mu = \frac{2}{3}\lambda \implies \mu = \frac{5}{3}\lambda\). For \(\mathbf{b}\): \(\frac{3}{5}\mu = 1 - \lambda\). Substituting the first into the second: \(\frac{3}{5}\left(\frac{5}{3}\lambda\right) = 1 - \lambda \implies \lambda = 1 - \lambda \implies \lambda = \frac{1}{2}\). Thus \mu = \frac{5}{6}\). Substitute \(\mu = \frac{5}{6}\) back into \(\overrightarrow{OX}\) to get \(\overrightarrow{OX} = \frac{5}{6}\left(\frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\right) = \frac{1}{3}\mathbf{a} + \frac{1}{2}\mathbf{b}\).

M1: For finding \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\) and \(\overrightarrow{OQ} = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\).
M1: For expressing \(\overrightarrow{OX}\) as \(\mu\left(\frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\right)\) or equivalent.
M1: For expressing \(\overrightarrow{OX}\) as \(\mathbf{b} + \lambda\left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right)\) or equivalent.
M1: For setting up simultaneous equations by equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\), and solving for \(\lambda\) or \(\mu\).
A1: For \(\overrightarrow{OX} = \frac{1}{3}\mathbf{a} + \frac{1}{2}\mathbf{b}\).

Compute the determinant of \(\mathbf{M}\): \(\det \mathbf{M} = (2a)(a-2) - 3(a+1) = 2a^2 - 4a - 3a - 3 = 2a^2 - 7a - 3\). Set the determinant equal to \(-9\): \(2a^2 - 7a - 3 = -9 \implies 2a^2 - 7a + 6 = 0\). Factorise the quadratic equation: \((2a - 3)(a - 2) = 0\). This gives the two values: \(a = 1.5\) and \(a = 2\). For part (b), the integer value of \(a\) is \(2\). Substituting \(a = 2\) into \(\mathbf{M}\): \(\mathbf{M} = \begin{pmatrix} 4 & 3 \\ 3 & 0 \end{pmatrix}\). Since \(\det \mathbf{M} = -9\), the inverse is: \(\mathbf{M}^{-1} = \frac{1}{-9} \begin{pmatrix} 0 & -3 \\ -3 & 4 \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{3} \\ \frac{1}{3} & -\frac{4}{9} \end{pmatrix}\).

M1: For writing the determinant equation: \(2a(a-2) - 3(a+1) = -9\).
A1: For simplifying to the correct quadratic equation: \(2a^2 - 7a + 6 = 0\).
M1: For solving the quadratic equation to get \(a = 1.5\) and \(a = 2\).
M1: For substituting the integer value \(a = 2\) into \(\mathbf{M}\) and applying the inverse matrix formula, e.g., \(\frac{1}{-9}\begin{pmatrix} 0 & -3 \\ -3 & 4 \end{pmatrix}\).
A1: For the correct inverse matrix \(\begin{pmatrix} 0 & \frac{1}{3} \\ \frac{1}{3} & -\frac{4}{9} \end{pmatrix}\) (or equivalent).

(a) The total number of balls in the bag is \(n + 4\). The probability of choosing two red balls without replacement is \(\frac{n}{n+4} \times \frac{n-1}{n+3} = \frac{1}{3}\). Multiplying both sides by \(3(n+4)(n+3)\) yields \(3n(n-1) = (n+4)(n+3)\). Expanding the brackets gives \(3n^2 - 3n = n^2 + 7n + 12\). Rearranging all terms to one side gives \(2n^2 - 10n - 12 = 0\). Dividing by 2 gives the required equation: \(n^2 - 5n - 6 = 0\). (b) Factorising the equation \((n-6)(n+1) = 0\) gives \(n = 6\) (since \(n > 0\)). The total number of balls is \(6 + 4 = 10\). The probability of selecting different colours is \(P(\text{Red then Blue}) + P(\text{Blue then Red}) = \left(\frac{6}{10} \times \frac{4}{9}\right) + \left(\frac{4}{10} \times \frac{6}{9}\right) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}\).

M1: For setting up the initial probability equation: \(\frac{n}{n+4} \times \frac{n-1}{n+3} = \frac{1}{3}\).
M1: For expanding and simplifying to show the given quadratic: \(3n(n-1) = (n+4)(n+3) \implies n^2 - 5n - 6 = 0\).
B1: For identifying \(n = 6\) (rejecting \(n = -1\)).
M1: For a correct expression for the probability of different colours using \(n=6\), e.g., \(2 \times \frac{6}{10} \times \frac{4}{9}\).
A1: For \(\frac{8}{15}\) (or equivalent fraction or decimal \(\approx 0.533\)).

(a) The area of a triangle is given by \(\text{Area} = \frac{1}{2} ac \sin B\). Substituting the known values: \(28 = \frac{1}{2} \times 10 \times 7 \times \sin(\angle ABC) \implies 28 = 35 \sin(\angle ABC) \implies \sin(\angle ABC) = 0.8\). The acute angle with a sine of \(0.8\) is \(\approx 53.13^\circ\). Since angle \(ABC\) is obtuse, \(\angle ABC = 180^\circ - 53.13^\circ = 126.87^\circ \approx 126.9^\circ\). (b) Apply the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle ABC) \implies AC^2 = 7^2 + 10^2 - 2 \times 7 \times 10 \times \cos(126.87^\circ)\). Since \(\cos(126.87^\circ) = -0.6\), we have \(AC^2 = 49 + 100 - 140 \times (-0.6) = 149 + 84 = 233\). Thus, \(AC = \sqrt{233} \approx 15.3\text{ cm}\).

M1: For using the area of a triangle formula: \(\frac{1}{2} \times 7 \times 10 \times \sin(\angle ABC) = 28\).
A1: For \(\angle ABC = 126.9^\circ\) (or \(126.87^\circ\) or better).
M1: For applying the cosine rule to find \(AC\), e.g., \(AC^2 = 7^2 + 10^2 - 2(7)(10)\cos(\text{their } \angle ABC)\).
A1: For correctly substituting values into the cosine rule, yielding \(AC^2 = 233\) (or value between \(232.8\) and \(233.2\)).
A1: For \(AC = 15.3\text{ cm}\) (must be rounded to 3 significant figures).

(a) A matrix does not have an inverse if its determinant is zero. \( \det(\mathbf{A}) = k(k+3) - 2(-1) = k^2 + 3k + 2 \). Setting this to zero: \( k^2 + 3k + 2 = 0 \implies (k+1)(k+2) = 0 \), so \( k = -1 \) or \( k = -2 \). (b) For \( k = 2 \), \( \mathbf{A} = \begin{pmatrix} 2 & 2 \\ -1 & 5 \end{pmatrix} \). The determinant is \( \det(\mathbf{A}) = 2(5) - 2(-1) = 12 \). The adjugate matrix is \( \begin{pmatrix} 5 & -2 \\ 1 & 2 \end{pmatrix} \). Thus, \( \mathbf{A}^{-1} = \frac{1}{12} \begin{pmatrix} 5 & -2 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{5}{12} & -\frac{1}{6} \\ \frac{1}{12} & \frac{1}{6} \end{pmatrix} \). (c) The simultaneous equations can be written in matrix form as \( \mathbf{A} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 8 \\ -10 \end{pmatrix} \). Solving for the variables: \( \begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{A}^{-1} \begin{pmatrix} 8 \\ -10 \end{pmatrix} = \frac{1}{12} \begin{pmatrix} 5 & -2 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 8 \\ -10 \end{pmatrix} = \frac{1}{12} \begin{pmatrix} 40 + 20 \\ 8 - 20 \end{pmatrix} = \frac{1}{12} \begin{pmatrix} 60 \\ -12 \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \end{pmatrix} \). So, \( x = 5 \) and \( y = -1 \).

Part (a): M1 for setting up the determinant expression and equating to 0. A1 for the quadratic equation \( k^2 + 3k + 2 = 0 \). A1 for both values: \( k = -1 \) and \( k = -2 \). Part (b): M1 for finding the determinant as 12 or finding the adjugate matrix. A2 for the fully correct inverse matrix (deduct 1 mark for any minor arithmetic error). Part (c): M1 for multiplying \( \mathbf{A}^{-1} \) by the column matrix. A1 for \( x = 5 \). A1 for \( y = -1 \).

(a) Time for the first part of the journey is \( \frac{60}{x} \) hours. Time for the second part is \( \frac{60}{x-10} \) hours. The total time is 5 hours: \( \frac{60}{x} + \frac{60}{x-10} = 5 \). Dividing the entire equation by 5 gives: \( \frac{12}{x} + \frac{12}{x-10} = 1 \). Multiplying by the common denominator \( x(x-10) \) yields: \( 12(x-10) + 12x = x(x-10) \implies 12x - 120 + 12x = x^2 - 10x \implies 24x - 120 = x^2 - 10x \). Rearranging into standard quadratic form: \( x^2 - 34x + 120 = 0 \). (b) Factoring the quadratic equation: \( (x-30)(x-4) = 0 \), which gives \( x = 30 \) or \( x = 4 \). Since the speed of the second part of the journey is \( x - 10 \) km/h, \( x \) must be greater than 10 for the speed to be positive. Therefore, \( x = 4 \) is rejected as a physical speed, leaving \( x = 30 \).

Part (a): M1 for writing expressions for time: \( \frac{60}{x} \) and \( \frac{60}{x-10} \). M1 for the equation \( \frac{60}{x} + \frac{60}{x-10} = 5 \). M1 for clearing denominators: \( 60(x-10) + 60x = 5x(x-10) \). A1 for intermediate simplification. A1 for obtaining exactly \( x^2 - 34x + 120 = 0 \) with no errors shown. Part (b): M1 for factoring or using the quadratic formula: \( (x-30)(x-4) = 0 \). A1 for obtaining both solutions \( x = 30 \) and \( x = 4 \). M1 for the explanation that speed must be positive so \( x > 10 \). A1 for stating \( x = 30 \) as the final unique answer.

(a) Let \( y = \frac{2x+3}{x-2} \). Multiplying both sides by \( x-2 \): \( y(x-2) = 2x+3 \implies yx - 2y = 2x + 3 \). Grouping terms in \( x \): \( yx - 2x = 2y + 3 \implies x(y-2) = 2y + 3 \implies x = \frac{2y+3}{y-2} \). Thus, \( \mathrm{f}^{-1}(x) = \frac{2x+3}{x-2} \). The domain of \( \mathrm{f}^{-1} \) is the range of \( \mathrm{f} \), which excludes the horizontal asymptote \( y = 2 \). Hence, the domain of \( \mathrm{f}^{-1} \) is \( x \in \mathbb{R}, x \neq 2 \). (b) The composite function is \( \mathrm{fg}(x) = \mathrm{f}(3x-1) = \frac{2(3x-1)+3}{(3x-1)-2} = \frac{6x-2+3}{3x-3} = \frac{6x+1}{3x-3} \). Setting this equal to 4: \( \frac{6x+1}{3x-3} = 4 \implies 6x+1 = 4(3x-3) \implies 6x+1 = 12x-12 \implies 6x = 13 \implies x = \frac{13}{6} \).

Part (a): M1 for setting up the equation \( y = \frac{2x+3}{x-2} \) and multiplying by \( x-2 \). M1 for collecting terms to isolate \( x \). A1 for the correct inverse formula \( \mathrm{f}^{-1}(x) = \frac{2x+3}{x-2} \). A1 for specifying the correct domain \( x \neq 2 \). Part (b): M1 for the substitution of \( \mathrm{g}(x) \) into \( \mathrm{f} \). A1 for the simplified expression \( \frac{6x+1}{3x-3} \). M1 for equating to 4 and multiplying by the denominator. A1 for a correct linear equation such as \( 6x = 13 \). A1 for \( x = \frac{13}{6} \) (or equivalent decimal \( 2.17 \) to 3 s.f.).

(a)(i) \( \overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a} \). (ii) Since \( AQ : QB = 3 : 2 \), \( \overrightarrow{AQ} = \frac{3}{5}\overrightarrow{AB} = \frac{3}{5}(\mathbf{b} - \mathbf{a}) \). Then \( \overrightarrow{OQ} = \overrightarrow{OA} + \overrightarrow{AQ} = \mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a}) = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b} \). (b) Since \( OP : PA = 2 : 1 \), \( \overrightarrow{OP} = \frac{2}{3}\mathbf{a} \). We find \( \overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \left(\frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\right) - \frac{2}{3}\mathbf{a} = -\frac{4}{15}\mathbf{a} + \frac{3}{5}\mathbf{b} \). Now, \( \overrightarrow{OR} = \overrightarrow{OQ} + \overrightarrow{QR} = \overrightarrow{OQ} + k\overrightarrow{PQ} = \left(\frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\right) + k\left(-\frac{4}{15}\mathbf{a} + \frac{3}{5}\mathbf{b}\right) = \left(\frac{2}{5} - \frac{4}{15}k\right)\mathbf{a} + \left(\frac{3}{5} + \frac{3}{5}k\right)\mathbf{b} \). Since \( R \) lies on \( OB \), the coefficient of \( \mathbf{a} \) must be 0: \( \frac{2}{5} - \frac{4}{15}k = 0 \implies k = \frac{2}{5} \times \frac{15}{4} = \frac{3}{2} = 1.5 \). Substituting \( k = 1.5 \) back into the expression for \( \overrightarrow{OR} \): \( \overrightarrow{OR} = \left(\frac{3}{5} + \frac{3}{5} \times \frac{3}{2}\right)\mathbf{b} = \left(\frac{6}{10} + \frac{9}{10}\right)\mathbf{b} = \frac{15}{10}\mathbf{b} = 1.5\mathbf{b} \).

Part (a)(i): B1 for \( \mathbf{b} - \mathbf{a} \). Part (a)(ii): M1 for the path \( \overrightarrow{OA} + \frac{3}{5}\overrightarrow{AB} \) or equivalent. A1 for \( \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b} \). Part (b)(i): M1 for expressing \( \overrightarrow{PQ} = -\frac{4}{15}\mathbf{a} + \frac{3}{5}\mathbf{b} \). M1 for writing \( \overrightarrow{OR} = \overrightarrow{OQ} + k\overrightarrow{PQ} \) and expanding. M1 for setting the coefficient of \( \mathbf{a} \) in \( \overrightarrow{OR} \) to 0. A1 for \( k = 1.5 \). Part (b)(ii): A1 for \( 1.5\mathbf{b} \) (or equivalent fraction).

(a) The angle \( \angle BAC \) is the difference between the two bearings: \( \angle BAC = 135^\circ - 60^\circ = 75^\circ \). In triangle \( ABC \), we have \( AB = 12 \), \( AC = 18 \), and the included angle \( \angle BAC = 75^\circ \). Applying the Cosine Rule to find \( BC \): \( BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC) \implies BC^2 = 12^2 + 18^2 - 2(12)(18)\cos(75^\circ) \implies BC^2 = 144 + 324 - 432(0.258819) = 356.190 \implies BC = \sqrt{356.190} \approx 18.873 \approx 18.9 \text{ km} \). (b) Using the Sine Rule to find the angle \( \angle ABC \): \( \frac{\sin(\angle ABC)}{18} = \frac{\sin(75^\circ)}{18.873} \implies \sin(\angle ABC) = \frac{18 \sin(75^\circ)}{18.873} \approx 0.92123 \implies \angle ABC \approx 67.11^\circ \). Since the bearing of \( B \) from \( A \) is \( 060^\circ \), the bearing of \( A \) from \( B \) is \( 180^\circ + 60^\circ = 240^\circ \). Since \( C \) is situated to the south-east of \( B \), the bearing of \( C \) from \( B \) is calculated by rotating counterclockwise from \( BA \) by the angle \( \angle ABC \): \( 240^\circ - 67.11^\circ = 172.89^\circ \approx 173^\circ \).

Part (a): M1 for determining angle \( \angle BAC = 75^\circ \). M1 for substituting correctly into the Cosine Rule. A1 for \( BC^2 \approx 356.2 \). A1 for \( BC \approx 18.9 \text{ km} \) (accept range 18.8 to 19.0). Part (b): M1 for applying the Sine Rule (or Cosine Rule) to find angle \( \angle ABC \). A1 for obtaining \( \angle ABC \approx 67.1^\circ \) (or 67^\circ). M1 for establishing the bearing of \( A \) from \( B \) is \( 240^\circ \) (or using parallel line angles). M1 for the subtraction step: \( 240^\circ - 67.11^\circ \). A1 for the final bearing of \( 173^\circ \) (accept range 172.5 to 173.5).

(a) The probability of drawing a red marble first is \( \frac{5}{n} \). Without replacement, the probability of drawing a red marble second is \( \frac{4}{n-1} \). The probability that both marbles are red is \( \frac{5}{n} \times \frac{4}{n-1} = \frac{20}{n(n-1)} \). (b)(i) Equating the expression to \( \frac{5}{33} \): \( \frac{20}{n(n-1)} = \frac{5}{33} \). Divide both sides by 5: \( \frac{4}{n(n-1)} = \frac{1}{33} \implies n(n-1) = 132 \implies n^2 - n - 132 = 0 \). (ii) Factoring the quadratic: \( (n-12)(n+11) = 0 \). Since \( n \) must be a positive integer, we reject \( n = -11 \) and obtain \( n = 12 \). (c) The probability that at least one marble is blue is the complement of the probability that both marbles are red: \( \mathrm{P}(\text{at least one Blue}) = 1 - \mathrm{P}(\text{both Red}) = 1 - \frac{5}{33} = \frac{28}{33} \).

Part (a): M1 for the product of two probabilities with denominators \( n \) and \( n-1 \). A1 for \( \frac{20}{n(n-1)} \). Part (b)(i): M1 for setting \( \frac{20}{n(n-1)} = \frac{5}{33} \). A1 for algebraic steps leading to \( n^2 - n - 132 = 0 \) without any gaps. Part (b)(ii): M1 for factoring to \( (n-12)(n+11) \) or using the quadratic formula. A1 for \( n = 12 \). Part (c): M1 for \( 1 - \mathrm{P}(\text{both Red}) \) or summing the individual probabilities of Red-Blue, Blue-Red, and Blue-Blue. A1 for substituting values. A1 for \( \frac{28}{33} \) (accept 0.848 or better).

(a) Let the three intersecting sets be labeled \( P \), \( J \), and \( C \). Starting from the innermost region: \( P \cap J \cap C = x \). The two-language-only regions are: \( P \cap J \cap C' = 15-x \), \( J \cap C \cap P' = 16-x \), and \( P \cap C \cap J' = 14-x \). The single-language-only regions are: Only \( P = 40 - (15-x + 14-x + x) = 11+x \); Only \( J = 45 - (15-x + 16-x + x) = 14+x \); Only \( C = 48 - (14-x + 16-x + x) = 18+x \). Outside the three sets is 8. (b) The total number of surveyed students is 100. Summing all the regions: \( (11+x) + (14+x) + (18+x) + (15-x) + (16-x) + (14-x) + x + 8 = 100 \implies 96 + x = 100 \implies x = 4 \). (c) The number of students who used Python or Java, but not C++ is: \( \text{Only } P + \text{Only } J + P \cap J \cap C' = (11+x) + (14+x) + (15-x) = 40 + x \). Substituting \( x = 4 \): \( 40 + 4 = 44 \).

Part (a): M1 for drawing three intersecting circles. A1 for labeling the three intersection-only regions: \( 15-x \), \( 16-x \), \( 14-x \). A1 for the single-language-only regions: \( 11+x \), \( 14+x \), \( 18+x \). A1 for the outside value of 8 and correct set labels. Part (b): M1 for setting up the equation representing the total sum of all regions equal to 100. A1 for simplifying the equation to \( 96 + x = 100 \). A1 for \( x = 4 \). Part (c): M1 for identifying the correct regions to sum: Only \( P \) + Only \( J \) + \( (P \cap J) \text{ only} \). A1 for obtaining 44.

(a) Total volume is the sum of the volume of the cylinder and the hemisphere: \( V = \pi r^2 h + \frac{2}{3} \pi r^3 = 120\pi \). Dividing by \( \pi \) gives: \( r^2 h + \frac{2}{3} r^3 = 120 \implies r^2 h = 120 - \frac{2}{3} r^3 \implies h = \frac{120}{r^2} - \frac{2}{3} r \). (b) Total surface area of the toy comprises the cylinder base, the cylinder curved surface, and the hemisphere curved surface: \( A = \pi r^2 + 2\pi r h + 2\pi r^2 = 3\pi r^2 + 2\pi r h \). Substitute the expression for \( h \): \( A = 3\pi r^2 + 2\pi r \left( \frac{120}{r^2} - \frac{2}{3} r \right) = 3\pi r^2 + \frac{240\pi}{r} - \frac{4}{3}\pi r^2 = \frac{5}{3}\pi r^2 + \frac{240\pi}{r} \). (c) Setting \( A = 95\pi \): \( \frac{5}{3}\pi r^2 + \frac{240\pi}{r} = 95\pi \implies \frac{5}{3} r^2 + \frac{240}{r} = 95 \). Multiply both sides by \( 3r \): \( 5r^3 + 720 = 285r \implies 5r^3 - 285r + 720 = 0 \). Dividing by 5: \( r^3 - 57r + 144 = 0 \). Since \( r \) is a positive integer, checking factors of 144: for \( r = 3 \), \( 3^3 - 57(3) + 144 = 27 - 171 + 144 = 0 \). Thus, \( r = 3 \).

Part (a): M1 for the formula \( \pi r^2 h + \frac{2}{3} \pi r^3 = 120\pi \). M1 for isolating the \( h \) term. A1 for \( h = \frac{120}{r^2} - \frac{2}{3} r \). Part (b): M1 for the formula for surface area: \( A = 3\pi r^2 + 2\pi r h \). M1 for substituting the expression for \( h \). A1 for algebraic simplification leading to \( A = \frac{5}{3}\pi r^2 + \frac{240\pi}{r} \) without errors. Part (c): M1 for setting up the equation \( \frac{5}{3} r^2 + \frac{240}{r} = 95 \). M1 for forming the cubic equation \( r^3 - 57r + 144 = 0 \). A1 for obtaining \( r = 3 \).

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

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

(c) We express \(\overrightarrow{OR}\) in two ways:
1) \(\overrightarrow{OR} = \mu \overrightarrow{OQ} = \frac{1}{2}\mu\mathbf{a} + \frac{1}{2}\mu\mathbf{b}\)
2) \(\overrightarrow{OR} = \overrightarrow{OB} + \overrightarrow{BR} = \mathbf{b} + \lambda\overrightarrow{BP} = \mathbf{b} + \lambda\left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\lambda\mathbf{a} + (1 - \lambda)\mathbf{b}\)

Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):
\(\frac{1}{2}\mu = \frac{2}{3}\lambda\) (Equation 1)
\(\frac{1}{2}\mu = 1 - \lambda\) (Equation 2)

Substitute Equation 1 into Equation 2:
\ \frac{2}{3}\lambda = 1 - \lambda \implies \frac{5}{3}\lambda = 1 \implies \lambda = \frac{3}{5}\)

Substitute \(\lambda = \frac{3}{5}\) back into Equation 1:
\(\mu = \frac{4}{3}\lambda = \frac{4}{3} \times \frac{3}{5} = \frac{4}{5}\).

(d) Substitute \(\mu = \frac{4}{5}\) into the expression for \(\overrightarrow{OR}\):
\(\overrightarrow{OR} = \frac{4}{5}\left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = \frac{2}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\).

(a) M1 for recognizing \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\) or writing a valid path for \(\overrightarrow{BP}\) such as \(\overrightarrow{BO} + \overrightarrow{OP}\). A1 for \(\frac{2}{3}\mathbf{a} - \mathbf{b}\).
(b) B1 for \(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\) or any equivalent simplified form.
(c) M1 for expressing \(\overrightarrow{OR}\) in terms of \(\lambda\) as \(\frac{2}{3}\lambda\mathbf{a} + (1 - \lambda)\mathbf{b}\). M1 for setting up two simultaneous equations by equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\). A1 for \(\lambda = \frac{3}{5}\). A1 for \(\mu = \frac{4}{5}\).
(d) B1.1 for \(\overrightarrow{OR} = \frac{2}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\) (or equivalent simplified vector expression).

(a) Time = \(\frac{\text{Distance}}{\text{Speed}}\), so the original time is \(\frac{36}{x}\) hours.

(b) For the slower journey, the speed is \((x - 3)\) km/h, so the time is \(\frac{36}{x - 3}\) hours.

(c) The slower journey takes 1 hour longer than the original journey:
\(\frac{36}{x - 3} - \frac{36}{x} = 1\)

Multiply both sides of the equation by \(x(x - 3)\) to clear the denominators:
\(36x - 36(x - 3) = x(x - 3)\)

Expand both sides:
\(36x - 36x + 108 = x^2 - 3x\)

Simplify:
\(108 = x^2 - 3x \implies x^2 - 3x - 108 = 0\).

(d) Factorize the quadratic equation:
\(x^2 - 3x - 108 = 0\)
\((x - 12)(x + 9) = 0\)

So, \(x = 12\) or \(x = -9\).

Since the speed \(x\) must be positive, we reject \(x = -9\), so \(x = 12\) km/h.

The time taken for the original journey is:
\(\text{Time} = \frac{36}{12} = 3\) hours.

(a) B1 for \(\frac{36}{x}\).
(b) B1 for \(\frac{36}{x - 3}\).
(c) M1 for forming the correct algebraic fraction equation: \(\frac{36}{x - 3} - \frac{36}{x} = 1\). M1 for multiplying both sides by \(x(x - 3)\) to obtain \(36x - 36(x - 3) = x(x - 3)\). A1 for expanding correctly to \(36x - 36x + 108 = x^2 - 3x\). A1 for fully convincing algebraic completion to show \(x^2 - 3x - 108 = 0\).
(d) M1 for attempt to solve the quadratic equation by factorization or quadratic formula, yielding at least one correct factor or root. A1 for \(x = 12\) (with positive speed context realized). A1.1 for finding the original time of 3 hours.

(a) In the right-angled triangle \(TFA\), \(\tan(32^\circ) = \frac{h}{AF}\), which gives \(AF = \frac{h}{\tan(32^\circ)}\).

(b) In the right-angled triangle \(TFB\), \(\tan(43^\circ) = \frac{h}{BF}\), which gives \(BF = \frac{h}{\tan(43^\circ)}\).

(c) Since \(A\) is due South of \(F\) and \(B\) is due East of \(F\), the angle \(\angle AFB = 90^\circ\). Hence, triangle \(AFB\) is right-angled at \(F\).
By Pythagoras' Theorem:
\(AF^2 + BF^2 = AB^2\)

Substitute the expressions from (a) and (b):
\(\left(\frac{h}{\tan(32^\circ)}\right)^2 + \left(\frac{h}{\tan(43^\circ)}\right)^2 = 25^2\)

\(\frac{h^2}{\tan^2(32^\circ)} + \frac{h^2}{\tan^2(43^\circ)} = 625\)

Factoring out \(h^2\):
\(h^2 \left( \frac{1}{\tan^2(32^\circ)} + \frac{1}{\tan^2(43^\circ)} \right) = 625\).

(d) Evaluate the trigonometric terms:
\(\tan(32^\circ) \approx 0.624869 \implies \tan^2(32^\circ) \approx 0.390462 \implies \frac{1}{\tan^2(32^\circ)} \approx 2.56107\)
\(\tan(43^\circ) \approx 0.932515 \implies \tan^2(43^\circ) \approx 0.869584 \implies \frac{1}{\tan^2(43^\circ)} \approx 1.14997\)

Substitute these back into the equation:
\(h^2 (2.56107 + 1.14997) = 625\)
\(h^2 (3.71104) = 625\)
\(h^2 = \frac{625}{3.71104} \approx 168.416\)
\(h = \sqrt{168.416} \approx 12.977\)

To 3 significant figures, \(h = 13.0\) meters.

(a) B1 for \(AF = \frac{h}{\tan(32^\circ)}\).
(b) B1 for \(BF = \frac{h}{\tan(43^\circ)}\).
(c) M1 for establishing \(AF^2 + BF^2 = AB^2\) or \(AF^2 + BF^2 = 25^2\) using Pythagoras' Theorem. M1 for substituting the expressions from (a) and (b) into the Pythagoras equation. A1 for factoring out \(h^2\) to show the given identity clearly.
(d) M1 for evaluating \(\frac{1}{\tan^2(32^\circ)}\) and \(\frac{1}{\tan^2(43^\circ)}\) (or evaluating \(AF\) and \(BF\) in terms of \(h\)). M1 for simplifying the sum inside the bracket to \(3.71\) (or equivalent accuracy). M1 for finding \(h^2 \approx 168.4\). A1.1 for \(13.0\) (accept \(13\), but must show 3 s.f. rounding from \(12.98\)).