2025 Edexcel IGCSE Mathematics (Specification B) 模擬試題連答案詳解

\(\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\)
\(A = \{2, 3, 5, 7, 11\}\)
\(B = \{3, 6, 9, 12\}\)
\(A \cup B = \{2, 3, 5, 6, 7, 9, 11, 12\}\)

The complement of \(A \cup B\) is \((A \cup B)' = \{1, 4, 8, 10\}\).

Therefore, \(n((A \cup B)') = 4\).

M1 for listing elements of \(A \cup B\) or elements of \((A \cup B)'\) with at most one omission/error.
A1 for correctly listing \((A \cup B)' = \{1, 4, 8, 10\}\).
A1 for final answer of 4.

Multiply both sides of the inequality by 12 to clear the denominators:
\(4(2x - 5) < 3(3x + 1)\)

Expand the brackets:
\(8x - 20 < 9x + 3\)

Rearrange to solve for \(x\):
\(-20 - 3 < 9x - 8x\)
\(-23 < x\)

which can be written as:
\(x > -23\).

M1 for multiplying by a common denominator to clear fractions, e.g., \(4(2x-5) < 3(3x+1)\).
M1 for correct expansion and gathering of like terms, e.g., \(8x - 20 < 9x + 3\).
A1 for the correct final inequality \(x > -23\) (or \(-23 < x\)).

First, write the term inside the bracket using positive indices:
\(\frac{8a^6}{27b^{-3}} = \frac{8a^6 b^3}{27}\)

Now apply the negative exponent by taking the reciprocal:
\(\left(\frac{8a^6 b^3}{27}\right)^{-\frac{1}{3}} = \left(\frac{27}{8a^6 b^3}\right)^{\frac{1}{3}}\)

Apply the power of \(\frac{1}{3}\) (cube root) to both the numerator and denominator:
\(\frac{27^{\frac{1}{3}}}{(8a^6 b^3)^{\frac{1}{3}}} = \frac{3}{2a^2 b\).

M1 for dealing with the negative index or the \(b^{-3}\) term correctly, e.g., obtaining \(\left(\frac{27}{8a^6 b^3}\right)^{\frac{1}{3}}\).
M1 for taking the cube root of the coefficients, i.e., finding \(\frac{3}{2}\) or \(1.5\).
A1 for the fully simplified expression \(\frac{3}{2a^2b}\) (accept \(1.5a^{-2}b^{-1}\)).

Form the composite function \(\mathrm{fg}(x)\):
\(\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x)) = \mathrm{f}\left(\frac{4}{x+1}\right) = 3\left(\frac{4}{x+1}\right) - 2\)

Set \(\mathrm{fg}(x) = 1\):
\(3\left(\frac{4}{x+1}\right) - 2 = 1\)
\(\frac{12}{x+1} = 3\)

Multiply by \(x + 1\):
\(12 = 3(x + 1)\)
\(12 = 3x + 3\)
\(3x = 9\)
\(x = 3\).

M1 for substituting \(\mathrm{g}(x)\) into \(\mathrm{f}(x)\) to get \(3\left(\frac{4}{x+1}\right) - 2\).
M1 for setting their composite expression equal to 1 and attempting to solve for \(x\), e.g., leading to \(\frac{12}{x+1} = 3\).
A1 for \(x = 3\).

The sum of an interior angle and an exterior angle of a regular polygon is \(180^\circ\).
Therefore, the size of each exterior angle is:
\(180^\circ - 162^\circ = 18^\circ\)

The sum of the exterior angles of any polygon is \(360^\circ\).
So, the number of sides \(n\) is:
\(n = \frac{360^\circ}{18^\circ} = 20\).

M1 for calculating the exterior angle: \(180^\circ - 162^\circ = 18^\circ\), or setting up the interior angle equation \(\frac{(n-2) \times 180}{n} = 162\).
M1 for attempting to divide 360 by their exterior angle, or solving their algebraic equation for \(n\).
A1 for 20.

Given that the ratio of blue to yellow counters is \(2 : 5\), and there are 15 yellow counters:
Number of blue counters \(= 15 \times \frac{2}{5} = 6\).

Total number of blue and yellow counters \(= 15 + 6 = 21\).

The probability of picking a blue or yellow counter is \(1 - P(\text{red}) = 1 - 0.3 = 0.7\).

Therefore, the 21 blue and yellow counters represent \(70\%\) of the total counters.
Let \(T\) be the total number of counters:
\(0.7T = 21\)
\(T = \frac{21}{0.7} = 30\).

M1 for finding the number of blue counters as 6 or the total blue + yellow as 21.
M1 for identifying that the combined probability of blue and yellow is \(0.7\).
A1 for the total number of counters being 30.

The formula for the total surface area \(A\) of a solid cylinder is:
\(A = 2\pi r^2 + 2\pi r h\)

Substitute \(h = 3r\) into the formula:
\(A = 2\pi r^2 + 2\pi r(3r) = 2\pi r^2 + 6\pi r^2 = 8\pi r^2\)

We are given that \(A = 288\pi\), so:
\(8\pi r^2 = 288\pi\)
\(8r^2 = 288\)
\(r^2 = 36\)

Since \(r > 0\), we have:
\(r = 6\).

M1 for substituting \(h = 3r\) into the total surface area formula to get \(2\pi r^2 + 2\pi r(3r)\) (or \(8\pi r^2\)).
M1 for setting their expression for surface area equal to \(288\pi\) and solving for \(r^2\).
A1 for \(r = 6\).

First, find the vector \(\overrightarrow{AB}\):
\(\overrightarrow{AB} = \begin{pmatrix} 5 - 1 \\ 6 - (-2) \end{pmatrix} = \begin{pmatrix} 4 \\ 8 \end{pmatrix}\)

So, \(\mathbf{a} = 4\mathbf{i} + 8\mathbf{j}\).

Now, find the magnitude of \(\mathbf{a}\):
\(|\mathbf{a}| = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80}\)

Simplify the surd:
\(\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}\).

M1 for finding the vector \(\overrightarrow{AB} = \begin{pmatrix} 4 \\ 8 \end{pmatrix}\) (or finding difference in \(x\) and \(y\) coordinates as 4 and 8).
M1 for applying Pythagoras' theorem to find magnitude: \(\sqrt{4^2 + 8^2}\) or \(\sqrt{80}\).
A1 for the simplified surd \(4\sqrt{5}\).

First, list the elements of the universal set: \(\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\). List the elements of set \(A\) (factors of 24 in \(\mathcal{E}\)): \(A = \{1, 2, 3, 4, 6, 8, 12\}\). For set \(B\), solve the inequality: \(2x + 1 > 9 \Rightarrow 2x > 8 \Rightarrow x > 4\). So, \(B = \{5, 6, 7, 8, 9, 10, 11, 12\}\). The complement of \(B\) is \(B' = \{1, 2, 3, 4\}\). Now find the intersection of \(A\) and \(B'\): \(A \cap B' = \{1, 2, 3, 4\}\). Therefore, the number of elements in the intersection is \(\text{n}(A \cap B') = 4\).

M1 for correctly listing set \(A\) or identifying the condition \(x > 4\) for set \(B\). M1 for correctly listing \(B' = \{1, 2, 3, 4\}\) or finding the set \(A \cap B'\). A1 for 4.

Factorise each quadratic expression: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\), \(4x^2 - 1 = (2x - 1)(2x + 1)\), \(2x^2 + 7x + 3 = (2x + 1)(x + 3)\), and \(x^2 - 9 = (x - 3)(x + 3)\). Substitute these factorised forms back into the expression: \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} \times \frac{(2x + 1)(x + 3)}{(x - 3)(x + 3)}\). Cancel out common terms in the numerator and denominator (namely \(2x+1\), \(x-3\), and \(x+3\)) to get: \(\frac{2x + 1}{2x - 1}\).

M1 for factorising at least two of the quadratic expressions correctly. M1 for fully factorising all four expressions. A1 for the correct simplified fraction \(\frac{2x + 1}{2x - 1}\).

We can use the property of determinants: \(\det(\mathbf{AB}) = \det(\mathbf{A}) \times \det(\mathbf{B})\). First, find \(\det(\mathbf{B}) = (2)(3) - (-1)(4) = 6 + 4 = 10\). Next, find \(\det(\mathbf{A}) = (2)(3) - (k)(-1) = 6 + k\). Using the determinant property: \((6 + k) \times 10 = 80 \Rightarrow 6 + k = 8 \Rightarrow k = 2\). Alternatively, calculate \(\mathbf{AB} = \begin{pmatrix} 4+4k & 3k-2 \\ 10 & 10 \end{pmatrix}\), so \(\det(\mathbf{AB}) = 10(4+4k) - 10(3k-2) = 10k + 60 = 80 \Rightarrow k = 2\).

M1 for finding \(\det(\mathbf{B}) = 10\) or expressing \(\mathbf{AB}\) correctly in terms of \(k\). M1 for setting up the equation \(10(6+k) = 80\) or \(10k + 60 = 80\). A1 for \(k = 2\).

Let \(P\) be the profits in 2021. The change from 2021 to 2022 is a multiplier of 1.15. The change from 2022 to 2023 is a multiplier of 0.92. Therefore, \(P \times 1.15 \times 0.92 = 243340 \Rightarrow P \times 1.058 = 243340\). Solving for \(P\): \(P = \frac{243340}{1.058} = 230000\).

M1 for expressing the multi-step percentages as multipliers: \(1.15\) and \(0.92\). M1 for setting up the equation \(P \times 1.15 \times 0.92 = 243340\). A1 for 230000 (accept $230,000).

The area of a triangle is given by \(\text{Area} = \frac{1}{2} a c \sin B\). Here, \(33 = \frac{1}{2} (11)(8) \sin B \Rightarrow 33 = 44 \sin B \Rightarrow \sin B = 0.75\). Since angle \(ABC\) is obtuse, \(B = 180^\circ - \arcsin(0.75) \approx 131.41^\circ\). Now use the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos B \Rightarrow AC^2 = 8^2 + 11^2 - 2(8)(11) \cos(131.41^\circ) \Rightarrow AC^2 = 64 + 121 - 176(-0.6614) \approx 185 + 116.41 = 301.41\). Taking the square root, \(AC \approx 17.36\text{ cm}\). To 3 significant figures, this is 17.4.

M1 for \(33 = \frac{1}{2} \times 8 \times 11 \times \sin B\) and finding \(\sin B = 0.75\). M1 for calculating the obtuse angle \(B \approx 131.4^\circ\) or using \(\cos B = -\sqrt{1 - 0.75^2} \approx -0.6614\). M1 for substituting values into the Cosine Rule: \(AC^2 = 8^2 + 11^2 - 2(8)(11)\cos(131.4^\circ)\). A1 for 17.4 (accept 17.3 to 17.4).

First, find the vector \(\mathbf{PQ}\): \(\mathbf{PQ} = \mathbf{OQ} - \mathbf{OP} = \begin{pmatrix} 15 \\ 7 \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 12 \\ 8 \end{pmatrix}\). Since \(PR : RQ = 3 : 1\), we have \(\mathbf{PR} = \frac{3}{4} \mathbf{PQ} = \frac{3}{4} \begin{pmatrix} 12 \\ 8 \end{pmatrix} = \begin{pmatrix} 9 \\ 6 \end{pmatrix}\). The position vector of \(R\) is \(\mathbf{OR} = \mathbf{OP} + \mathbf{PR} = \begin{pmatrix} 3 \\ -1 \end{pmatrix} + \begin{pmatrix} 9 \\ 6 \end{pmatrix} = \begin{pmatrix} 12 \\ 5 \end{pmatrix}\). The magnitude of \(\mathbf{OR}\) is \(|\mathbf{OR}| = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\).

M1 for finding \(\mathbf{PQ} = \begin{pmatrix} 12 \\ 8 \end{pmatrix}\). M1 for finding \(\mathbf{OR} = \begin{pmatrix} 12 \\ 5 \end{pmatrix}\). A1 for 13.

The volume of a hemisphere is \(V = \frac{2}{3} \pi R^3\). For a radius of 6 cm, \(V = \frac{2}{3} \pi (6)^3 = 144\pi\text{ cm}^3\). This volume equals the volume of the cone, \(V = \frac{1}{3} \pi r^2 h\), where \(r = 4\text{ cm}\). Thus, \(144\pi = \frac{1}{3} \pi (4)^2 h \Rightarrow 144 = \frac{16}{3} h \Rightarrow h = \frac{144 \times 3}{16} = 27\text{ cm}\). The slant height \(l\) of the cone is \(l = \sqrt{r^2 + h^2} = \sqrt{4^2 + 27^2} = \sqrt{16 + 729} = \sqrt{745} \approx 27.295\text{ cm}\). The total surface area of the cone is \(A = \pi r^2 + \pi r l = \pi (4)^2 + \pi (4)(\sqrt{745}) = 16\pi + 4\sqrt{745}\pi \approx 50.27 + 343.00 = 393.27\text{ cm}^2\). To 3 significant figures, this is 393.

M1 for equating the volume of the hemisphere to the volume of the cone: \(\frac{2}{3} \pi (6)^3 = \frac{1}{3} \pi (4)^2 h\) and finding \(h = 27\). M1 for calculating slant height \(l = \sqrt{4^2 + 27^2} \approx 27.3\). M1 for calculating total surface area \(\pi (4)^2 + \pi (4)(27.295)\). A1 for 393 (accept 393 to 394).

The total number of counters is 20. The number of ways to choose 3 counters from 20 is \(\binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140\). We want the probability of getting at least two red counters, which means either exactly 2 red and 1 blue, or exactly 3 red. The number of ways to choose exactly 2 red and 1 blue is \(\binom{12}{2} \times \binom{8}{1} = 66 \times 8 = 528\). The number of ways to choose exactly 3 red is \(\binom{12}{3} = \frac{12 \times 11 \times 10}{6} = 220\). The total number of successful outcomes is \(528 + 220 = 748\). The probability is \(\frac{748}{1140} = \frac{187}{285}\).

M1 for finding total outcomes \(1140\) or a correct probability tree path calculation. M1 for calculating the number of ways for 2 red and 1 blue (528) OR 3 red (220), or equivalent probabilities: \(3 \times \frac{12}{20} \times \frac{11}{19} \times \frac{8}{18}\) or \(\frac{12}{20} \times \frac{11}{19} \times \frac{10}{18}\). M1 for adding the two appropriate cases. A1 for \(\frac{187}{285}\) (or equivalent simplified fraction).

The universal set is \( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\} \). The set of prime numbers is \( A = \{2, 3, 5, 7, 11, 13\} \). The set of multiples of 3 is \( B = \{3, 6, 9, 12, 15\} \). The union of \( A \) and \( B \) is \( A \cup B = \{2, 3, 5, 6, 7, 9, 11, 12, 13, 15\} \). There are 10 elements in \( A \cup B \). Therefore, the complement \( (A \cup B)' \) contains the remaining elements \( \{1, 4, 8, 10, 14\} \). Thus, \( \text{n}((A \cup B)') = 5 \).

M1: For listing the elements of \( A \) and \( B \) or showing \( A \cup B \) has 10 elements. M1: For identifying the elements of the complement set \( (A \cup B)' = \{1, 4, 8, 10, 14\} \) or subtracting 10 from 15. A1: For 5.

Let the profits at the start of 2021 be \( P \). After a 15% increase, the profits are \( 1.15P \). After an 8% decrease, the profits are \( 1.15P \times 0.92 = 1.058P \). We are given that \( 1.058P = 126960 \). Solving for \( P \) gives \( P = \frac{126960}{1.058} = 120000 \). Therefore, the profits at the start of 2021 were £120,000.

M1: For expressing the net multiplier as \( 1.15 \times 0.92 = 1.058 \). M1: For establishing the equation \( 1.058P = 126960 \) or the division \( 126960 / 1.058 \). A1: For 120000 (accept £120,000).

First, factorise each quadratic and cubic expression: \( 2x^2 + 5x - 3 = (2x - 1)(x + 3) \), \( 4x^2 - 1 = (2x - 1)(2x + 1) \), and \( x^3 + 3x^2 = x^2(x + 3) \). Now substitute these into the expression: \( \frac{(2x - 1)(x + 3)}{(2x - 1)(2x + 1)} \div \frac{x^2(x + 3)}{2x + 1} \). Simplify the first fraction by cancelling \( (2x - 1) \): \( \frac{x + 3}{2x + 1} \div \frac{x^2(x + 3)}{2x + 1} \). Multiply by the reciprocal of the second fraction: \( \frac{x + 3}{2x + 1} \times \frac{2x + 1}{x^2(x + 3)} \). Cancel common terms to get: \( \frac{1}{x^2} \).

M1: For factorising \( 2x^2 + 5x - 3 \) as \( (2x-1)(x+3) \) or \( 4x^2 - 1 \) as \( (2x-1)(2x+1) \). M1: For factorising \( x^3 + 3x^2 \) as \( x^2(x+3) \) and multiplying by the reciprocal of the divisor. A1: For the final answer \( \frac{1}{x^2} \) or \( x^{-2} \).

The composite function is \( \text{fg}(x) = \text{f}(\text{g}(x)) = 3\left(\frac{4}{x+1}\right) - 2 \). Setting this equal to 1, we get: \( 3\left(\frac{4}{x+1}\right) - 2 = 1 \Rightarrow 3\left(\frac{4}{x+1}\right) = 3 \Rightarrow \frac{4}{x+1} = 1 \Rightarrow x + 1 = 4 \Rightarrow x = 3 \).

M1: For setting up the equation \( 3\left(\frac{4}{x+1}\right) - 2 = 1 \) or establishing \( \text{g}(x) = \text{f}^{-1}(1) \). M1: For isolating \( \frac{4}{x+1} = 1 \) or equivalent linear step. A1: For 3.

By the intersecting chords theorem, we have \( AP \times PB = CP \times PD \). Substituting the given values: \( 6(x + 2) = 4(2x - 1) \). Expanding both sides: \( 6x + 12 = 8x - 4 \). Subtracting \( 6x \) and adding 4 on both sides: \( 2x = 16 \Rightarrow x = 8 \).

M1: For writing \( 6(x + 2) = 4(2x - 1) \) or stating the intersecting chords relation. M1: For expanding and rearranging to a linear form \( 2x = 16 \). A1: For 8.

The arc length of the sector is equal to the base circumference of the cone. Arc length = \( 2 \pi (12) \times \frac{150}{360} = 10\pi\text{ cm} \). Let \( r \) be the base radius of the cone, then \( 2\pi r = 10\pi \Rightarrow r = 5\text{ cm} \). The slant height \( l \) of the cone is equal to the radius of the sector, so \( l = 12\text{ cm} \). The perpendicular height \( h \) of the cone is \( h = \sqrt{l^2 - r^2} = \sqrt{12^2 - 5^2} = \sqrt{119} \approx 10.9\text{ cm} \) (to 3 s.f.).

M1: For finding the base radius of the cone \( r = 5 \) or the arc length of the sector \( 10\pi \). M1: For applying Pythagoras' theorem to find the perpendicular height: \( \sqrt{12^2 - 5^2} \). A1: For 10.9.

We can write \( \vec{PQ} = \vec{PA} + \vec{AQ} \). Since \( OP : PA = 2 : 1 \), we have \( \vec{PA} = \frac{1}{3}\vec{OA} = \frac{1}{3}\mathbf{a} \). Also, \( \vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} \). Since \( AQ : QB = 3 : 2 \), we have \( \vec{AQ} = \frac{3}{5}\vec{AB} = \frac{3}{5}(\mathbf{b} - \mathbf{a}) \). Substituting these in: \( \vec{PQ} = \frac{1}{3}\mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a}) = \frac{1}{3}\mathbf{a} - \frac{3}{5}\mathbf{a} + \frac{3}{5}\mathbf{b} = -\frac{4}{15}\mathbf{a} + \frac{3}{5}\mathbf{b} \).

M1: For finding \( \vec{PA} = \frac{1}{3}\mathbf{a} \) or \( \vec{AQ} = \frac{3}{5}(\mathbf{b} - \mathbf{a}) \). M1: For combining vectors to express \( \vec{PQ} \) as a path of known vectors. A1: For \( \frac{3}{5}\mathbf{b} - \frac{4}{15}\mathbf{a} \) or equivalent simplest form.

The two outcomes that give different colours are Red then Blue (RB) and Blue then Red (BR). The total number of beads is 12. \( \text{P}(RB) = \frac{5}{12} \times \frac{7}{11} = \frac{35}{132} \). \( \text{P}(BR) = \frac{7}{12} \times \frac{5}{11} = \frac{35}{132} \). Thus, the total probability is \( \text{P}(\text{different}) = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66} \).

M1: For calculating the probability of one correct pair of different colours (e.g. \( \frac{5}{12} \times \frac{7}{11} \)). M1: For adding both valid combinations together: \( \text{P}(RB) + \text{P}(BR) \) with correct denominators. A1: For \( \frac{35}{66} \) or equivalent fraction (accept 0.530 or 53.0%).

To solve \(2^{3x - 1} \times 4^{x + 2} = 8^{2x - 3}\), express all terms as powers of the base 2: \(4 = 2^2\) and \(8 = 2^3\). Substitute these into the equation: \(2^{3x - 1} \times (2^2)^{x + 2} = (2^3)^{2x - 3}\). Apply the power of a power rule \((a^m)^n = a^{mn}\): \(2^{3x - 1} \times 2^{2(x + 2)} = 2^{3(2x - 3)}\), which simplifies to \(2^{3x - 1} \times 2^{2x + 4} = 2^{6x - 9}\). Apply the multiplication law of indices \(a^m \times a^n = a^{m+n}\) to the left-hand side: \(2^{(3x - 1) + (2x + 4)} = 2^{6x - 9}\), which simplifies to \(2^{5x + 3} = 2^{6x - 9}\). Equate the exponents: \(5x + 3 = 6x - 9\). Rearrange the terms to solve for \(x\): \(3 + 9 = 6x - 5x\), which gives \(x = 12\).

M1: For writing \(4\) as \(2^2\) and \(8\) as \(2^3\), or correctly rewriting at least one term (e.g., \(4^{x+2}\) as \(2^{2x+4}\) or \(8^{2x-3}\) as \(2^{6x-9}\)). M1: For combining the left-hand side into a single power of 2, yielding \(2^{5x+3}\). M1: For setting up the linear equation from the exponents: \(5x+3 = 6x-9\). A1: For the correct solution \(x = 12\).

Let \(N\) be the total number of students, so \(n(\mathcal{U}) = 60\). Let \(H\) be the set of students studying History and \(G\) be the set of students studying Geography. We are given \(n(H) = 35\), \(n(G) = 27\), and \(n((H \cup G)') = 12\). The number of students studying History or Geography or both is \(n(H \cup G) = 60 - 12 = 48\). Using the principle of inclusion-exclusion: \(n(H \cup G) = n(H) + n(G) - n(H \cap G)\). Substitute the known values: \(48 = 35 + 27 - n(H \cap G)\), which simplifies to \(48 = 62 - n(H \cap G)\). Solving for the intersection gives \(n(H \cap G) = 62 - 48 = 14\). So, 14 students study both subjects. The probability of choosing a student who studies both subjects at random is \(P(H \cap G) = \frac{n(H \cap G)}{n(\mathcal{U})} = \frac{14}{60}\). Simplifying this fraction to its simplest form gives \(\frac{7}{30}\).

M1: For calculating the number of students who study History or Geography (or both): \(60 - 12 = 48\). M1: For substituting into the union formula, or using a Venn diagram to form the equation: \(35 + 27 - x = 48\) (where \(x = n(H \cap G)\)). M1: For solving the equation to find \(n(H \cap G) = 14\). A1: For the correct fraction in its simplest form: \(\frac{7}{30}\).

(a)

The time taken by car A is \(\frac{36}{x}\) hours.

The time taken by car B is \(\frac{36}{x - 15}\) hours.

Since car B takes 12 minutes longer, and \(12\text{ minutes} = \frac{12}{60} = \frac{1}{5}\) hours, we can write:

\[\frac{36}{x - 15} - \frac{36}{x} = \frac{1}{5}\]

Multiply through by \(5x(x - 15)\):

\[5 \times 36x - 5 \times 36(x - 15) = x(x - 15)\]

\[180x - 180x + 2700 = x^2 - 15x\]

\[x^2 - 15x - 2700 = 0\]

(b)

We solve \(x^2 - 15x - 2700 = 0\) by factorisation:

\[(x - 60)(x + 45) = 0\]

Since speed must be positive, \(x = 60\).

The speed of car A is 60 km/h.

(c)

The speed of car B is \(60 - 15 = 45\) km/h.

The time taken by car B is \(\frac{36}{45}\text{ hours} = 0.8\text{ hours} = 48\text{ minutes}\).

Part (a):
- M1: Expressing times for both cars as \(\frac{36}{x}\) and \(\frac{36}{x-15}\).
- M1: Setting up the correct difference equation: \(\frac{36}{x-15} - \frac{36}{x} = \frac{12}{60}\) (or equivalent).
- M1: Eliminating denominators correctly by multiplying through by \(5x(x-15)\).
- A1: Showing clearly the transition to \(x^2 - 15x - 2700 = 0\) with no algebraic errors.

Part (b):
- M1: A method to solve the quadratic equation, either by factorisation \((x-60)(x+45)=0\) or using the quadratic formula.
- A1: Obtaining \(x = 60\) and rejecting \(x = -45\).

Part (c):
- M1: Finding the speed of car B (45 km/h) and setting up \(\frac{36}{45} \times 60\) or finding car A's time (36 mins) and adding 12.
- A1: 48.

From the first equation, express \(y\) in terms of \(x\):

\[y = x + 2\]

Substitute \(y = x + 2\) into the second equation:

\[2x^2 - x(x + 2) + (x + 2)^2 = 8\]

Expand the brackets:

\[2x^2 - (x^2 + 2x) + (x^2 + 4x + 4) = 8\]

\[2x^2 - x^2 - 2x + x^2 + 4x + 4 = 8\]

Combine like terms:

\[2x^2 + 2x + 4 = 8\]

\[2x^2 + 2x - 4 = 0\]

Divide the entire equation by 2:

\[x^2 + x - 2 = 0\]

Factorise the quadratic:

\[(x + 2)(x - 1) = 0\]

This gives two values for \(x\):

\[x = -2 \quad \text{or} \quad x = 1\]

Now substitute these values back into \(y = x + 2\) to find the corresponding \(y\) values:

- For \(x = -2\): \(y = -2 + 2 = 0\)
- For \(x = 1\): \(y = 1 + 2 = 3\)

Therefore, the solutions are \(x = -2, y = 0\) and \(x = 1, y = 3\).

- M1: Rearranging the linear equation to express one variable in terms of another (e.g., \(y = x + 2\)).
- M1: Substituting this expression into the quadratic equation.
- A1: Expanding brackets correctly to reach \(2x^2 - x^2 - 2x + x^2 + 4x + 4 = 8\) or equivalent.
- M1: Simplifying to a standard quadratic form, e.g., \(2x^2 + 2x - 4 = 0\) or \(x^2 + x - 2 = 0\).
- M1: Attempting to solve their quadratic equation by factorisation or formula.
- A1: Finding the two correct values of \(x\) (or \(y\)), i.e., \(x = -2\) and \(x = 1\).
- M1: Substituting both of their values back to find the corresponding other coordinates.
- A1: Fully correct paired solutions: \((-2, 0)\) and \((1, 3)\).

(a)

The value after 2 years is given by:

\[250000 \times \left(1 + \frac{x}{100}\right)^2 = 291600\]

\[\left(1 + \frac{x}{100}\right)^2 = \frac{291600}{250000} = 1.1664\]

\[1 + \frac{x}{100} = \sqrt{1.1664} = 1.08\]

\[\frac{x}{100} = 0.08 \implies x = 8\]

(b)

In 2022, the value is £291,600.

In 2023, the value increased by 8%:

\[V = 291600 \times 1.08 = 314928\]

In 2024, the value decreased by 12%:

\[\text{Value in 2024} = 314928 \times (1 - 0.12) = 314928 \times 0.88 = 277136.64\]

(c)

The overall percentage change from 2020 (£250,000) to 2024 (£277,136.64) is:

\[\text{Percentage Change} = \frac{277136.64 - 250000}{250000} \times 100 = \frac{27136.64}{250000} \times 100 \approx 10.854656\%\]

To 3 significant figures, this is a \(10.9\%\) increase.

Part (a):
- M1: Setting up the compound interest equation: \(250000 \times r^2 = 291600\).
- M1: Solving for \(r\) to find \(r = 1.08\).
- A1: Showing clearly that \(x = 8\).

Part (b):
- M1: Finding the value in 2023: \(291600 \times 1.08 = 314928\).
- M1: Applying the 12% decrease: \(314928 \times 0.88\).
- A1: £277,136.64 (accept £277,000 to 3 s.f.).

Part (c):
- M1: Using the formula \(\frac{\text{New} - \text{Old}}{\text{Old}} \times 100\).
- A1: Correctly calculating \(10.9\%\) (accept 10.85% or 10.8546...%).

(a)

Total counters = 12.

The probability of selecting two red counters is:

\[P(RR) = \frac{8}{12} \times \frac{7}{11} = \frac{56}{132}\]

The probability of selecting two blue counters is:

\[P(BB) = \frac{4}{12} \times \frac{3}{11} = \frac{12}{132}\]

\[P(\text{same colour}) = P(RR) + P(BB) = \frac{56}{132} + \frac{12}{132} = \frac{68}{132} = \frac{17}{33}\]

(b)

We need to find \(P(\text{3rd is Blue} \mid \text{1st and 2nd same colour})\).

By definition:

\[P(\text{3rd is Blue} \mid \text{same}) = \frac{P(\text{3rd is Blue and same})}{P(\text{same})}\]

The outcomes that satisfy "first two are the same colour AND the third is blue" are:

1) \(R, R, B\)

2) \(B, B, B\)

We compute these probabilities:

\[P(RRB) = \frac{8}{12} \times \frac{7}{11} \times \frac{4}{10} = \frac{224}{1320}\]

\[P(BBB) = \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} = \frac{24}{1320}\]

\[P(\text{3rd is Blue and same}) = \frac{224}{1320} + \frac{24}{1320} = \frac{248}{1320}\]

We also know:

\[P(\text{same}) = \frac{68}{132} = \frac{680}{1320}\]

Now, calculate the conditional probability:

\[P(\text{3rd is Blue} \mid \text{same}) = \frac{248/1320}{680/1320} = \frac{248}{680} = \frac{31}{85}\]

Part (a):
- M1: Correct calculation of \(P(RR) = \frac{8}{12} \times \frac{7}{11}\) or \(P(BB) = \frac{4}{12} \times \frac{3}{11}\).
- M1: Adding their two same-colour probabilities.
- A1: \(\frac{17}{33}\) (or equivalent fraction like \(\frac{68}{132}\), or decimal \(0.515\) to 3 s.f.).

Part (b):
- M1: Finding the probability of \(P(RRB) = \frac{8}{12} \times \frac{7}{11} \times \frac{4}{10}\).
- M1: Finding the probability of \(P(BBB) = \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10}\).
- M1: Adding their \(P(RRB)\) and \(P(BBB)\) to find the numerator.
- M1: Setting up the division by their \(P(\text{same colour})\) from part (a).
- A1: \(\frac{31}{85}\) (or equivalent fraction, or decimal \(0.365\) to 3 s.f.).

(a)

The total height of the solid is \(20\) cm. Since the hemisphere has radius \(r\) cm, the height of the cylinder is:

\[h = 20 - r\]

The volume of the solid is:

\[V = V_{\text{cylinder}} + V_{\text{hemisphere}}\]

\[V = \pi r^2 h + \frac{2}{3}\pi r^3\]

\[V = \pi r^2 (20 - r) + \frac{2}{3}\pi r^3\]

\[V = 20\pi r^2 - \pi r^3 + \frac{2}{3}\pi r^3 = 20\pi r^2 - \frac{1}{3}\pi r^3\]

(b)

Equating the volume to \(648\pi\):

\[20\pi r^2 - \frac{1}{3}\pi r^3 = 648\pi\]

Divide by \(\pi\):

\[20r^2 - \frac{1}{3}r^3 = 648\]

Multiply through by \(-3\):

\[r^3 - 60r^2 + 1944 = 0\]

(c)

Substitute \(r = 6\) into the equation:

\[6^3 - 60(6^2) + 1944 = 216 - 2160 + 1944 = 0\]

Since the result is 0, \(r = 6\) is a solution.

(d)

Since \(r = 6\) is a root, we can factorise \(r^3 - 60r^2 + 1944\) by dividing by \(r - 6\):

\[r^3 - 60r^2 + 1944 = (r - 6)(r^2 - 54r - 324) = 0\]

To find the other roots, solve \(r^2 - 54r - 324 = 0\) using the quadratic formula:

\[r = \frac{54 \pm \sqrt{(-54)^2 - 4(1)(-324)}}{2}\]

\[r = \frac{54 \pm \sqrt{2916 + 1296}}{2} = \frac{54 \pm \sqrt{4212}}{2}\]

\[r \approx \frac{54 \pm 64.8999}{2}\]

This gives:

\[r_1 \approx 59.5 \quad \text{and} \quad r_2 \approx -5.45\]

For this solid, the radius \(r\) must be positive and less than 20 cm (since the height of the cylinder \(h = 20 - r > 0\)).

- \(r = -5.45\) is negative and therefore not a possible physical length.
- \(r = 59.5\) is greater than 20, which would make the cylinder height negative.

Thus, only \(r = 6\) is a valid radius.

Part (a):
- M1: Expressing the height of the cylinder as \(20 - r\).
- M1: Setting up volume formula: \(V = \pi r^2 (20-r) + \frac{2}{3}\pi r^3\).
- A1: Simplifying to \(V = 20\pi r^2 - \frac{1}{3}\pi r^3\).

Part (b):
- M1: Equating their volume to \(648\pi\) and dividing both sides by \(\pi\).
- A1: Showing clear algebraic manipulation to achieve \(r^3 - 60r^2 + 1944 = 0\).

Part (c):
- B1: Substituting \(r = 6\) into the cubic equation and showing \(216 - 2160 + 1944 = 0\).

Part (d):
- M1: Dividing \(r^3 - 60r^2 + 1944\) by \(r - 6\) to find the quadratic factor \(r^2 - 54r - 324\).
- A1: Calculating the other two solutions: \(r \approx 59.5\) and \(r \approx -5.45\) (to 3 s.f.).
- A1: Providing a complete explanation for why \(r = 59.5\) and \(r = -5.45\) are rejected based on physical constraints of the solid.

(a)

We are given \(OP : PA = 2 : 1\), so \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).

Since \(AQ : QB = 3 : 2\), we can express \(\overrightarrow{AQ}\) as:

\[\overrightarrow{AQ} = \frac{3}{5}\overrightarrow{AB} = \frac{3}{5}(\mathbf{b} - \mathbf{a})\]

We find \(\overrightarrow{OQ}\):

\[\overrightarrow{OQ} = \overrightarrow{OA} + \overrightarrow{AQ} = \mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a}) = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\]

Now, express \(\overrightarrow{PQ}\):

\[\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \left(\frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\right) - \frac{2}{3}\mathbf{a} = \left(\frac{2}{5} - \frac{2}{3}\right)\mathbf{a} + \frac{3}{5}\mathbf{b} = -\frac{4}{15}\mathbf{a} + \frac{3}{5}\mathbf{b}\]

(b)

We have \(OB : BR = 1 : k\), so \(\overrightarrow{BR} = k\mathbf{b}\).

Thus, \(\overrightarrow{OR} = \overrightarrow{OB} + \overrightarrow{BR} = (1 + k)\mathbf{b}\).

Express \(\overrightarrow{PR}\):

\[\overrightarrow{PR} = \overrightarrow{OR} - \overrightarrow{OP} = (1 + k)\mathbf{b} - \frac{2}{3}\mathbf{a} = -\frac{2}{3}\mathbf{a} + (1 + k)\mathbf{b}\]

(c)

Since \(P\), \(Q\) and \(R\) lie on a straight line, \(\overrightarrow{PR}\) is a scalar multiple of \(\overrightarrow{PQ}\).

Let \(\overrightarrow{PR} = \lambda \overrightarrow{PQ}\).

\[-\frac{2}{3}\mathbf{a} + (1 + k)\mathbf{b} = \lambda \left(-\frac{4}{15}\mathbf{a} + \frac{3}{5}\mathbf{b}\right)\]

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

\[-\frac{2}{3} = -\frac{4}{15}\lambda \implies \lambda = \frac{2}{3} \times \frac{15}{4} = \frac{5}{2} = 2.5\]

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

\[1 + k = \frac{3}{5}\lambda = \frac{3}{5} \times \frac{5}{2} = \frac{3}{2}\]

\[k = \frac{3}{2} - 1 = \frac{1}{2} = 0.5\]

Part (a):
- M1: Finding \(\overrightarrow{OQ} = \mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a})\) (or equivalent path).
- M1: Setting up \(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}\) with \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).
- A1: \(-\frac{4}{15}\mathbf{a} + \frac{3}{5}\mathbf{b}\).

Part (b):
- M1: Finding \(\overrightarrow{OR} = (1 + k)\mathbf{b}\).
- A1: \(-\frac{2}{3}\mathbf{a} + (1 + k)\mathbf{b}\).

Part (c):
- M1: Using the collinearity condition \(\overrightarrow{PR} = \lambda \overrightarrow{PQ}\).
- M1: Equating coefficients of \(\mathbf{a}\) to find \(\lambda\).
- A1: Finding \(k = 0.5\) (or \(\frac{1}{2}\)).

(a)

The Venn diagram consists of three intersecting circles, \(F\), \(G\), and \(S\), surrounded by a universal set boundary.

- The central region \(F \cap G \cap S\) has \(x\).
- The region \(F \cap G \cap S'\) has 14.
- The region \(F \cap S \cap G'\) has 10.
- The region \(G \cap S \cap F'\) has 8.
- The French-only region has \(2x + 1\).
- The German-only region has \(x + 5\).
- The Spanish-only region has \(3x - 4\).
- The region outside all three circles has 16.

(b)

The sum of the elements in all regions must equal the total number of students, which is 120:

\[(2x + 1) + (x + 5) + (3x - 4) + 14 + 10 + 8 + x + 16 = 120\]

Group the \(x\) terms and constant terms:

\[(2x + x + 3x + x) + (1 + 5 - 4 + 14 + 10 + 8 + 16) = 120\]

\[7x + 50 = 120\]

\[7x = 70 \implies x = 10\]

(c)

Students who study at least two languages are in the regions representing intersections of two or more sets:

\[n(\text{at least 2}) = n(F \cap G \cap S') + n(F \cap S \cap G') + n(G \cap S \cap F') + n(F \cap G \cap S)\]

\[n(\text{at least 2}) = 14 + 10 + 8 + x\]

Since \(x = 10\):

\[n(\text{at least 2}) = 14 + 10 + 8 + 10 = 42\]

Part (a):
- M1: Correctly placing the intersection numbers \(14, 10, 8\) and variable \(x\) in their respective intersection regions.
- M1: Correctly placing \(2x + 1\), \(x + 5\), and \(3x - 4\) in the "only" regions.
- A1: Correctly placing 16 outside all three sets.
- A1: Fully correct Venn diagram with all labels.

Part (b):
- M1: Setting up the equation summing all regions equal to 120.
- A1: Simplifying to \(7x + 50 = 120\) and solving to get \(x = 10\).

Part (c):
- M1: Identifying the correct regions for "at least two languages" (\(14 + 10 + 8 + x\)).
- A1: 42.

(a)

Let \(y = f(x)\):

\[y = \frac{2x + 5}{x - 3}\]

Multiply both sides by \(x - 3\):

\[y(x - 3) = 2x + 5\]

\[xy - 3y = 2x + 5\]

Rearrange to group the terms with \(x\) on one side:

\[xy - 2x = 3y + 5\]

Factorise \(x\):

\[x(y - 2) = 3y + 5\]

\[x = \frac{3y + 5}{y - 2}\]

Thus, the inverse function is:

\[f^{-1}(x) = \frac{3x + 5}{x - 2}\]

(b)

We want to solve \(fg(x) = 3\).

This is equivalent to \(g(x) = f^{-1}(3)\).

First, find \(f^{-1}(3)\):

\[f^{-1}(3) = \frac{3(3) + 5}{3 - 2} = \frac{14}{1} = 14\]

Now, solve \(g(x) = 14\):

\[3x - 1 = 14\]

\[3x = 15 \implies x = 5\]

(c)

For the composite function \(fg(x) = f(g(x))\) to be defined, the output of \(g(x)\) must be within the domain of \(f(x)\).

Since \(f(x)\) is defined for \(x \neq 3\), we must have:

\[g(x) \neq 3\]

\[3x - 1 \neq 3\]

\[3x \neq 4 \implies x \neq \frac{4}{3}\]

Thus, the value \(x = \frac{4}{3}\) must be excluded from the domain of \(fg\).

Part (a):
- M1: Setting \(y = \frac{2x + 5}{x - 3}\) and multiplying by \((x - 3)\).
- M1: Grouping \(x\) terms together and factorising \(x(y - 2)\).
- A1: \(f^{-1}(x) = \frac{3x + 5}{x - 2}\).

Part (b):
- M1: Setting up the equation \(f(g(x)) = 3\) or stating \(g(x) = f^{-1}(3)\).
- M1: Finding \(f^{-1}(3) = 14\) or correctly expanding \(\frac{2(3x-1)+5}{(3x-1)-3} = 3\).
- A1: \(x = 5\).

Part (c):
- M1: Stating that the denominator of the composite function cannot be zero, i.e., \(g(x) \neq 3\).
- A1: \(x = 4/3\) (or \(1.33\) to 3 s.f.).

(a) Time taken for the outward journey is:
\[t_1 = \frac{150}{x}\]

(b) The return speed is \(x + 15\) km/h, so the time taken is:
\[t_2 = \frac{150}{x + 15}\]

(c) Since the return journey took 30 minutes (\(0.5\) hours) less than the outward journey:
\[\frac{150}{x} - \frac{150}{x + 15} = \frac{1}{2}\]

Multiply through by \(2x(x + 15)\) to clear the fractions:
\[300(x + 15) - 300x = x(x + 15)\]
\[300x + 4500 - 300x = x^2 + 15x\]
\[4500 = x^2 + 15x\]
\[x^2 + 15x - 4500 = 0\]

(d) We solve the quadratic equation:
\[(x - 60)(x + 75) = 0\]

This gives \(x = 60\) or \(x = -75\).
Since speed must be positive, we reject \(-75\), giving \(x = 60\).

(a)
B1: For \(\frac{150}{x}\) (or equivalent).

(b)
B1: For \(\frac{150}{x + 15}\) (or equivalent).

(c)
M1: For writing a correct equation relating the times, e.g., \(\frac{150}{x} - \frac{150}{x + 15} = 0.5\) or equivalent.
M1: For algebraic steps to clear the denominators, e.g., \(150(x+15) - 150x = 0.5x(x+15)\).
A1.33: For simplifying correctly to achieve the given quadratic equation: \(x^2 + 15x - 4500 = 0\) with no errors shown.

(d)
M1: For attempting to solve the quadratic equation by factoring to \((x - 60)(x + 75) = 0\) or using the quadratic formula.
A1: For finding the roots \(x = 60\) and \(x = -75\).
A1: For stating \(x = 60\) and clearly rejecting the negative root.

Let \(R_1, R_2\) represent drawing a red ball first and second, and \(B_1, B_2\) represent drawing a blue ball first and second.

(a) The probability that both balls are of the same colour is:
\[P(\text{Same}) = P(R_1 \cap R_2) + P(B_1 \cap B_2)\]
\[P(R_1 \cap R_2) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90}\]
\[P(B_1 \cap B_2) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90}\]
\[P(\text{Same}) = \frac{30}{90} + \frac{12}{90} = \frac{42}{90} = \frac{7}{15}\]

(b) The probability that the second ball is blue is:
\[P(B_2) = P(R_1 \cap B_2) + P(B_1 \cap B_2)\]
\[P(R_1 \cap B_2) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}\]
\[P(B_1 \cap B_2) = \frac{12}{90}\]
\[P(B_2) = \frac{24}{90} + \frac{12}{90} = \frac{36}{90} = \frac{2}{5}\]

(c) We want the conditional probability:
\[P(R_1 | B_2) = \frac{P(R_1 \cap B_2)}{P(B_2)}\]
\[P(R_1 | B_2) = \frac{24/90}{36/90} = \frac{24}{36} = \frac{2}{3}\]

(a)
M1: For writing or calculating \(P(R_1 \cap R_2) = \frac{6}{10} \times \frac{5}{9}\) or \(P(B_1 \cap B_2) = \frac{4}{10} \times \frac{3}{9}\).
M1: For adding their two correct joint probabilities.
A1: For \(\frac{7}{15}\) (or equivalent fraction/decimal, e.g., \(0.467\)).

(b)
M1: For attempting to sum \(P(R_1 \cap B_2)\) and \(P(B_1 \cap B_2)\).
A1.33: For \(\frac{2}{5}\) (or \(0.4\)).

(c)
M1: For using the conditional probability formula \(P(R_1 | B_2) = \frac{P(R_1 \cap B_2)}{P(B_2)}\).
M1: For substituting their values: \(\frac{24/90}{\text{their (b)}}\).
A1: For \(\frac{2}{3}\) (or equivalent fraction/decimal, e.g., \(0.667\)).

(a)
(i) \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}\)
(ii) Since \(Q\) is the midpoint of \(AB\):
\[\overrightarrow{OQ} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\]
(iii) Since \(OP : PA = 2 : 1\), we have \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).
\[\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) - \frac{2}{3}\mathbf{a} = -\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}\]

(b) We are given \(\overrightarrow{PR} = k\overrightarrow{PQ} = k\left(-\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}\right)\).
Thus:
\[\overrightarrow{OR} = \overrightarrow{OP} + \overrightarrow{PR} = \frac{2}{3}\mathbf{a} + k\left(-\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = \left(\frac{2}{3} - \frac{k}{6}\right)\mathbf{a} + \frac{k}{2}\mathbf{b}\]
Since \(R\) lies on the line \(OB\), the coefficient of \(\mathbf{a}\) must be 0:
\[\frac{2}{3} - \frac{k}{6} = 0 \implies \frac{k}{6} = \frac{2}{3} \implies k = 4\]

(c) Substituting \(k = 4\) into \(\overrightarrow{OR}\):
\[\overrightarrow{OR} = \frac{4}{2}\mathbf{b} = 2\mathbf{b}\]
Since \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OR} = 2\mathbf{b}\), we have:
\[\overrightarrow{BR} = \overrightarrow{OR} - \overrightarrow{OB} = 2\mathbf{b} - \mathbf{b} = \mathbf{b}\]
Thus, \(OB\) and \(BR\) have equal lengths, so the ratio \(OB : BR = 1 : 1\).

(a)
B1: For \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\).
B1: For \(\overrightarrow{OQ} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\) (or equivalent).
B1: For \(\overrightarrow{PQ} = -\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}\) (or equivalent).

(b)
M1: For expressing \(\overrightarrow{OR} = \overrightarrow{OP} + k\overrightarrow{PQ}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
M1: For equating the coefficient of \(\mathbf{a}\) to 0, i.e., \(\frac{2}{3} - \frac{k}{6} = 0\).
A1.33: For finding \(k = 4\).

(c)
M1: For finding \(\overrightarrow{OR} = 2\mathbf{b}\) or concluding \(\overrightarrow{BR} = \mathbf{b}\).
A1: For stating the ratio as \(1 : 1\) (or equivalent).

(a) The volume of a hemisphere of radius \(r\) is:
\[V_{\text{hemi}} = \frac{2}{3}\pi r^3\]
The volume of a cone of radius \(r\) and height \(h\) is:
\[V_{\text{cone}} = \frac{1}{3}\pi r^2 h\]
Since their volumes are equal:
\[\frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2 h\]
Dividing both sides by \(\frac{1}{3}\pi r^2\) (since \(r \neq 0\)) gives:
\[2r = h\]

(b) The total surface area of the toy consists of the curved surface area of the hemisphere and the curved surface area of the cone.
\[A = A_{\text{hemi}} + A_{\text{cone}} = 2\pi r^2 + \pi r l\]
where \(l\) is the slant height of the cone.
Using Pythagoras' theorem:
\[l = \sqrt{r^2 + h^2}\]
Since \(h = 2r\):
\[l = \sqrt{r^2 + (2r)^2} = \sqrt{5r^2} = r\sqrt{5}\]
Substitute \(l\) back into the surface area equation:
\[A = 2\pi r^2 + \pi r(r\sqrt{5}) = \pi r^2(2 + \sqrt{5})\]
We are given \(A = 180\pi\):
\[\pi r^2(2 + \sqrt{5}) = 180\pi\]
\[r^2(2 + \sqrt{5}) = 180\]
\[r^2 = \frac{180}{2 + \sqrt{5}}\]
\[r = \sqrt{\frac{180}{2 + \sqrt{5}}} \approx \sqrt{42.4922} \approx 6.5186\]
To 3 significant figures, \(r = 6.52\).

(a)
M1: For writing correct expressions for the volume of the hemisphere, \(\frac{2}{3}\pi r^3\), and the volume of the cone, \(\frac{1}{3}\pi r^2 h\).
A1: For setting them equal and correctly showing \(h = 2r\) with no errors.

(b)
M1: For writing the expression for the slant height of the cone, \(l = \sqrt{r^2 + h^2}\), and substituting \(h = 2r\) to find \(l = r\sqrt{5}\).
M1: For writing the expression for the total surface area of the toy, \(2\pi r^2 + \pi r l\).
M1: For substituting \(l = r\sqrt{5}\) into their surface area expression and equating to \(180\pi\).
M1: For solving for \(r^2\) to get \(r^2 = \frac{180}{2+\sqrt{5}}\) (or equivalent decimal value \(\approx 42.5\)).
A1.33: For calculating \(r \approx 6.52\) (accept range 6.51 - 6.53).