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

To simplify the expression, we first write each surd in its simplest form:
\(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\)
\(\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}\)
\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)

Substitute these back into the expression:
\(\frac{3(5\sqrt{3}) - 3\sqrt{3}}{2\sqrt{3}} = \frac{15\sqrt{3} - 3\sqrt{3}}{2\sqrt{3}}\)
\(\quad = \frac{12\sqrt{3}}{2\sqrt{3}}\)
\(\quad = 6\)

M1 for expressing at least two of \(\sqrt{75}\), \(\sqrt{27}\), or \(\sqrt{12}\) in the form \(k\sqrt{3}\) (e.g., \(5\sqrt{3}\), \(3\sqrt{3}\), or \(2\sqrt{3}\)).
M1 for simplifying the numerator to \(12\sqrt{3}\) or the whole fraction to \(\frac{12\sqrt{3}}{2\sqrt{3}}\).
A0.5 for \(6\).

Factorise the quadratic expression in the numerator:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)

Factorise the difference of two squares in the denominator:
\(4x^2 - 1 = (2x - 1)(2x + 1)\)

Substitute these factorised forms back into the fraction and cancel the common factor \((2x + 1)\):
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\)

M1 for factorising the numerator correctly as \((2x + 1)(x - 3)\).
M1 for factorising the denominator correctly as \((2x - 1)(2x + 1)\).
A0.5 for \(\frac{x - 3}{2x - 1}\).

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

Now, find \(B'\), which is the set of elements in \(\mathcal{U}\) that are not in \(B\):
\(B' = \{5, 7, 8, 9, 10, 11\}\)

Next, find the intersection \(A \cap B'\) (elements that are in both \(A\) and \(B'\)):
\(A \cap B' = \{5, 7, 11\}\)

Finally, find the number of elements in this intersection:
\(\text{n}(A \cap B') = 3\)

M1 for listing the elements of \(A\) and \(B\) correctly (at least one fully correct, or both mostly correct).
M1 for listing the elements of \(B'\) or identifying \(A \cap B' = \{5, 7, 11\}\).
A0.5 for \(3\).

To find \(\text{g}^{-1}(2)\), let \(\text{g}^{-1}(2) = x\), which is equivalent to \(\text{g}(x) = 2\).

Set up the equation:
\(\frac{3x - 1}{2x + 5} = 2\)

Multiply both sides by \(2x + 5\):
\(3x - 1 = 2(2x + 5)\)
\(3x - 1 = 4x + 10\)

Rearrange the terms to solve for \(x\):
\(-1 - 10 = 4x - 3x\)
\(x = -11\)

Thus, \(\text{g}^{-1}(2) = -11\).

M1 for setting up the equation \(\frac{3x - 1}{2x + 5} = 2\) or finding the algebraic inverse function expression \(\text{g}^{-1}(x) = \frac{-5x - 1}{2x - 3}\).
M1 for expanding and rearranging the equation to find \(x\) (or substituting \(x = 2\) into their inverse function expression).
A0.5 for \(-11\).

Since \(\mathbf{M}\) is a singular matrix, its determinant must be equal to 0:
\(\det(\mathbf{M}) = (k-2)(k) - (3 \times 8) = 0\)
\(k^2 - 2k - 24 = 0\)

Factorise the quadratic equation:
\((k - 6)(k + 4) = 0\)

This yields the solutions \(k = 6\) or \(k = -4\).
Since we are given that \(k > 0\), we reject the negative root.
Therefore, \(k = 6\).

M1 for setting up the determinant equation \((k-2)k - 24 = 0\).
M1 for factorising the quadratic equation to get \((k-6)(k+4) = 0\) or using the quadratic formula to obtain the roots.
A0.5 for \(6\) (must reject \(k = -4\)).

The total number of sweets in the bag is \(n + 5\).
The probability of selecting a red sweet first is \(\frac{5}{n+5}\).
Since the sweets are not replaced, the number of sweets remaining is \(n + 4\), of which 4 are red.
The probability of selecting a second red sweet is \(\frac{4}{n+4}\).

Set up the probability equation for both being red:
\(\frac{5}{n+5} \times \frac{4}{n+4} = \frac{2}{9}\)
\(\frac{20}{n^2 + 9n + 20} = \frac{2}{9}\)

Cross-multiply to solve:
\(20 \times 9 = 2(n^2 + 9n + 20)\)
\(180 = 2n^2 + 18n + 40\)

Divide by 2 and rearrange into standard quadratic form:
\(n^2 + 9n - 70 = 0\)

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

Since the number of sweets \(n\) must be positive, \(n = 5\).

M1 for writing the probability of two red sweets as \(\frac{5}{n+5} \times \frac{4}{n+4}\).
M1 for setting up the equation \(\frac{20}{(n+5)(n+4)} = \frac{2}{9}\) and reducing it to a standard quadratic equation of the form \(n^2 + 9n - 70 = 0\).
A0.5 for \(5\) (must reject \(n = -14\)).

First, calculate the magnitude of vector \(\mathbf{a}\):
\(|\mathbf{a}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\)

The magnitude of vector \(\mathbf{b}\) is 2.5 times this:
\(|\mathbf{b}| = 2.5 \times 10 = 25\)

Using the definition of magnitude for \(\mathbf{b}\):
\(|\mathbf{b}| = \sqrt{x^2 + 15^2} = 25\)

Square both sides of the equation:
\(x^2 + 225 = 625\)
\(x^2 = 400\)
\(x = \pm 20\)

Since we are given that \(x < 0\), the value of \(x\) must be \(-20\).

M1 for finding the magnitude of \(\mathbf{a}\) to be \(10\) or writing \(|\mathbf{b}| = 25\).
M1 for setting up the equation \(\sqrt{x^2 + 15^2} = 25\) and rearranging to get \(x^2 = 400\).
A0.5 for \(-20\) (must reject \(20\)).

The area of a triangle is given by the formula:
\(\text{Area} = \frac{1}{2} b c \sin(A)\)

Substitute the given values into the formula:
\(20\sqrt{3} = \frac{1}{2} \times 8 \times 10 \times \sin(BAC)\)
\(20\sqrt{3} = 40 \sin(BAC)\)

Solve for \(\sin(BAC)\):
\(\sin(BAC) = \frac{20\sqrt{3}}{40} = \frac{\sqrt{3}}{2}\)

The acute angle with a sine of \(\frac{\sqrt{3}}{2}\) is \(60^\circ\).
Since angle \(BAC\) is obtuse (between \(90^\circ\) and \(180^\circ\)):
\(BAC = 180^\circ - 60^\circ = 120^\circ\).

M1 for substituting the given values into the triangle area formula: \(\frac{1}{2} \times 8 \times 10 \times \sin(BAC) = 20\sqrt{3}\).
M1 for isolating the sine value: \(\sin(BAC) = \frac{\sqrt{3}}{2}\) or \(\sin(BAC) \approx 0.866\).
A0.5 for \(120\) (or \(120^\circ\)).

First, we list the elements of the universal set and the subsets: \(\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\), \(A = \{3, 6, 9, 12\}\), and \(B = \{1, 2, 3, 4, 6, 8, 12\}\). Next, we find the complement of \(A\): \(A' = \{1, 2, 4, 5, 7, 8, 10, 11\}\). Now, we find the intersection of \(A'\) and \(B\): \(A' \cap B = \{1, 2, 4, 8\}\). Therefore, the number of elements is \(\text{n}(A' \cap B) = 4\).

M1: For listing the elements of \(B\) correctly or finding the set \(A'\) correctly. A1: For 4.

We first calculate \(\mathbf{M}^2\): \(\mathbf{M}^2 = \begin{pmatrix} 2 & k \\ -1 & 3 \end{pmatrix} \begin{pmatrix} 2 & k \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 2(2) + k(-1) & 2(k) + k(3) \\ -1(2) + 3(-1) & -1(k) + 3(3) \end{pmatrix} = \begin{pmatrix} 4 - k & 5k \\ -5 & 9 - k \end{pmatrix}\). By comparing \(\mathbf{M}^2\) with the given matrix \(\begin{pmatrix} 2 & 10 \\ -5 & 7 \end{pmatrix}\), we have: \(4 - k = 2 \implies k = 2\), \(5k = 10 \implies k = 2\), and \(9 - k = 7 \implies k = 2\). Thus, the value of the constant \(k\) is 2.

M1: Attempt to multiply matrix \(\mathbf{M}\) by itself to get at least two correct algebraic elements. M1: Set up a correct linear equation in \(k\) using corresponding elements. A1: For \(k = 2\).

First, find the composite function \(\mathrm{fg}(x)\): \(\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x)) = 2(x^2 + 1) - 3 = 2x^2 + 2 - 3 = 2x^2 - 1\). Now set \(\mathrm{fg}(x) = 17\): \(2x^2 - 1 = 17 \implies 2x^2 = 18 \implies x^2 = 9\). Solving for \(x\) gives \(x = 3\) or \(x = -3\).

M1: For finding the correct expression for \(\mathrm{fg}(x)\) as \(2x^2 - 1\). M1: For setting \(2x^2 - 1 = 17\) and solving for \(x^2\). A1: For both \(x = 3\) and \(x = -3\) (or write \(x = \pm 3\)).

Rationalising the denominator of the first term: \(\frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}\). Simplifying the second term: \(\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}\). Adding the two terms together: \(4\sqrt{3} + 3\sqrt{3} = 7\sqrt{3}\). Therefore, \(a = 7\).

M1: For rationalising to get \(4\sqrt{3}\) or simplifying to get \(3\sqrt{3}\). A1: For \(a = 7\).

First, find the sum of the vectors \(\mathbf{a} + \mathbf{b}\): \(\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 + 12 \\ -4 + p \end{pmatrix} = \begin{pmatrix} 15 \\ p - 4 \end{pmatrix}\). The magnitude of this vector is given as 17: \(\sqrt{15^2 + (p - 4)^2} = 17\). Squaring both sides: \(225 + (p - 4)^2 = 289 \implies (p - 4)^2 = 64\). Taking the square root: \(p - 4 = 8\) or \(p - 4 = -8\). This gives \(p = 12\) or \(p = -4\). Since we are given that \(p > 0\), we reject \(p = -4\), leaving \(p = 12\).

M1: For expressing \(\mathbf{a} + \mathbf{b}\) as \(\begin{pmatrix} 15 \\ p - 4 \end{pmatrix}\). M1: For setting up the equation \(15^2 + (p - 4)^2 = 17^2\) and attempting to solve for \(p\). A1: For \(p = 12\) (must reject \(p = -4\)).

Multiply both sides by \((2 - y)\): \(x(2 - y) = 3y + 4 \implies 2x - xy = 3y + 4\). Rearrange to bring all terms containing \(y\) to one side: \(2x - 4 = 3y + xy\). Factor out \(y\) on the right-hand side: \(2x - 4 = y(3 + x)\). Divide by \((3 + x)\) to make \(y\) the subject: \(y = \frac{2x - 4}{x + 3}\).

M1: For clear intention to multiply by the denominator and isolate the terms containing \(y\) on one side. A1: For \(y = \frac{2x - 4}{x + 3}\) or any equivalent expression such as \(y = \frac{4 - 2x}{-x - 3}\).

Let the sum of the initial 8 numbers be \(S_8\). Since the mean of these 8 numbers is 15, we have: \(\frac{S_8}{8} = 15 \implies S_8 = 8 \times 15 = 120\). Let the sum of the 9 numbers be \(S_9\). Since the new mean is 17, we have: \(\frac{S_9}{9} = 17 \implies S_9 = 9 \times 17 = 153\). The 9th number is the difference between these two sums: \(S_9 - S_8 = 153 - 120 = 33\).

M1: For finding the sum of the first 8 numbers (120). M1: For finding the sum of the 9 numbers (153). A1: For 33.

In the right-angled triangle \(ABC\) with the right angle at \(B\), the side \(AB\) is opposite to angle \(ACB\) and \(AC\) is the hypotenuse. Using the sine ratio: \(\sin(\angle ACB) = \frac{AB}{AC} = \frac{7}{12}\). To find angle \(ACB\): \(\angle ACB = \arcsin\left(\frac{7}{12}\right) \approx 35.685^\circ\). Rounding to 1 decimal place gives \(35.7^\circ\).

M1: For identifying and using the sine ratio, i.e., \(\sin(\angle ACB) = \frac{7}{12}\). M1: For finding the inverse sine of \(\frac{7}{12}\). A1: For 35.7 (accept 35.68 or better).

First, multiply both sides by the denominator: \(y(2x + 5) = 7 - 3x\). Next, expand the left side: \(2xy + 5y = 7 - 3x\). Group all terms containing \(x\) on one side and terms without \(x\) on the other side: \(2xy + 3x = 7 - 5y\). Factorise \(x\) on the left side: \(x(2y + 3) = 7 - 5y\). Finally, divide by the bracket to make \(x\) the subject: \(x = \dfrac{7 - 5y}{2y + 3}\).

M1 for multiplying by \(2x + 5\) and expanding to obtain \(2xy + 5y = 7 - 3x\). M1 for collecting terms in \(x\) on one side and factorising to obtain \(x(2y + 3) = 7 - 5y\) (or equivalent). A1 for the final correct expression: \(x = \dfrac{7 - 5y}{2y + 3}\) (or equivalent, such as \(x = \dfrac{5y - 7}{-2y - 3}\)).

We use the property that \(\det(\mathbf{A}\mathbf{B}) = \det(\mathbf{A}) \times \det(\mathbf{B})\). First, find the determinant of \(\mathbf{A}\): \(\det(\mathbf{A}) = 2(4) - k(-3) = 8 + 3k\). Next, find the determinant of \(\mathbf{B}\): \(\det(\mathbf{B}) = 5(-1) - 1(2) = -7\). Now, equate the product to \(-77\): \((8 + 3k)(-7) = -77\). Divide both sides by \(-7\): \(8 + 3k = 11\) which simplifies to \(3k = 3\), so \(k = 1\).

M1: For finding an expression for \(\det(\mathbf{A}) = 8 + 3k\). M1: For calculating \(\det(\mathbf{B}) = -7\). M1: For setting up the equation \((8 + 3k)(-7) = -77\). A1: For simplifying to \(8 + 3k = 11\). A1: For solving to find \(k = 1\).

From the first equation, we get \(y = 2x - 3\). Substituting this into the second equation gives: \(x^2 + (2x - 3)^2 = 5\). Expanding the squared term: \(x^2 + 4x^2 - 12x + 9 = 5\). Simplifying yields: \(5x^2 - 12x + 4 = 0\). Factoring the quadratic: \((5x - 2)(x - 2) = 0\). This gives \(x = 2\) or \(x = 0.4\). When \(x = 2\), \(y = 2(2) - 3 = 1\). When \(x = 0.4\), \(y = 2(0.4) - 3 = -2.2\).

M1: Expressing \(y\) in terms of \(x\) as \(y = 2x - 3\). M1: Substituting into the quadratic equation to form an equation in one variable. M1: Expanding and simplifying to \(5x^2 - 12x + 4 = 0\). M1: Solving the quadratic equation. A1: Correct \(x\) values: \(x = 2\) and \(x = 0.4\). A1: Correct paired \(y\) values: \(y = 1\) and \(y = -2.2\).

The total number of students in the union of the three sets is \(100 - 12 = 88\). Using the principle of inclusion-exclusion: \(88 = 45 + 40 + 35 - (15 + 10 + 12) + x\), which simplifies to \(88 = 120 - 37 + x\), so \(88 = 83 + x\) and thus \(x = 5\). The number of students studying only Biology is \(45 - (10 - 5 + 5 + 5) = 45 - 20 = 25\). The number of students studying only Chemistry is \(40 - (10 + 5 + 7) = 18\). The number of students studying only Physics is \(35 - (5 + 5 + 7) = 18\). The total number of students who study exactly one subject is \(25 + 18 + 18 = 61\).

M1: Finding the total number of students in the union is 88. M1: Setting up the inclusion-exclusion equation to find \(x\). A1: Solving for \(x = 5\). M1: Calculating the number of students studying only one subject for at least one subject. A1: Correct values for Only B (25), Only C (18), and Only P (18). A1: Summing the values to get 61.

Let \(A\) be the event Bag A is chosen and \(B\) be the event Bag B is chosen. \(P(A) = 2/6 = 1/3\) and \(P(B) = 4/6 = 2/3\). Let \(R\) be the event of selecting a red ball. The conditional probabilities are \(P(R|A) = 4/10 = 2/5\) and \(P(R|B) = 5/8\). The total probability of selecting a red ball is \(P(R) = P(A)P(R|A) + P(B)P(R|B) = (1/3)(2/5) + (2/3)(5/8) = 2/15 + 10/24 = 8/60 + 25/60 = 33/60 = 11/20\). Using Bayes' theorem, \(P(B|R) = P(B \cap R) / P(R) = (5/12) / (11/20) = (5/12) * (20/11) = 100/132 = 25/33\).

M1: Identifying \(P(A) = 1/3\) and \(P(B) = 2/3\). M1: Writing the correct total probability formula for red. M1: Calculating individual branch probabilities: \(2/15\) and \(5/12\). A1: Correct total probability \(P(R) = 11/20\) (or 0.55). M1: Applying the formula for conditional probability. A1: Correct simplified fraction 25/33.

To find \(\mathrm{f}^{-1}(x)\), let \(y = \frac{3x+2}{x-4}\). Multiplying both sides by \(x-4\) gives \(y(x-4) = 3x+2 \implies xy - 4y = 3x+2 \implies xy - 3x = 4y+2 \implies x(y-3) = 4y+2 \implies x = \frac{4y+2}{y-3}\). Thus, \(\mathrm{f}^{-1}(x) = \frac{4x+2}{x-3}\). Setting \(\mathrm{f}^{-1}(x) = \mathrm{g}(x)\) gives \(\frac{4x+2}{x-3} = 2x+1 \implies 4x+2 = (2x+1)(x-3) \implies 4x+2 = 2x^2 - 5x - 3 \implies 2x^2 - 9x - 5 = 0\). Factoring this quadratic gives \((2x+1)(x-5) = 0\). This yields the solutions \(x = 5\) and \(x = -0.5\).

M1: Rearranging the equation to make \(x\) the subject. A1: Correct inverse function \(\mathrm{f}^{-1}(x) = \frac{4x+2}{x-3}\). M1: Setting the inverse equal to \(\mathrm{g}(x)\) and clearing the fraction. A1: Correct quadratic equation \(2x^2 - 9x - 5 = 0\). M1: Solving the quadratic equation. A1: Finding both correct solutions \(x = 5\) and \(x = -0.5\).

We rewrite the equation using indices: \(3(3^x)^2 - 10(3^x) + 3 = 0\). Let \(y = 3^x\), which gives the quadratic equation: \(3y^2 - 10y + 3 = 0\). Factoring the quadratic gives: \((3y - 1)(y - 3) = 0\), so \(y = 1/3\) or \(y = 3\). Replacing \(y\) with \(3^x\): \(3^x = 1/3 \implies x = -1\) and \(3^x = 3 \implies x = 1\).

M1: Rewriting \(3^{2x+1}\) as \(3 \cdot (3^x)^2\). M1: Substituting \(y = 3^x\) to obtain a quadratic. A1: Finding the correct quadratic equation \(3y^2 - 10y + 3 = 0\). M1: Solving the quadratic equation to get \(y = 1/3\) and \(y = 3\). A1: Finding \(3^x = 1/3\) and \(3^x = 3\). A1: Finding the final correct answers \(x = -1\) and \(x = 1\).

Using the area formula: \(\text{Area} = \frac{1}{2} a c \sin B\), we have \(14\sqrt{5} = \frac{1}{2} (9)(7) \sin(\angle ABC) \implies 14\sqrt{5} = \frac{63}{2} \sin(\angle ABC) \implies \sin(\angle ABC) = \frac{28\sqrt{5}}{63} = \frac{4\sqrt{5}}{9}\). Since \(\angle ABC\) is acute, \(\cos(\angle ABC) = \sqrt{1 - \sin^2(\angle ABC)} = \sqrt{1 - \frac{80}{81}} = \frac{1}{9}\). Using the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC) = 7^2 + 9^2 - 2(7)(9)\left(\frac{1}{9}\right) = 49 + 81 - 14 = 116\). Thus, \(AC = \sqrt{116} = 2\sqrt{29}\).

M1: Using the area of a triangle formula with given values. M1: Calculating the exact value of \(\sin(\angle ABC) = \frac{4\sqrt{5}}{9}\). M1: Finding the exact value of \(\cos(\angle ABC) = \frac{1}{9}\). M1: Correct application of the Cosine Rule to find \(AC^2\). A1: Finding \(AC^2 = 116\). A1: Simplifying to \(2\sqrt{29}\).

First, \(\overrightarrow{OP} = \mathbf{a} + \frac{3}{4}\overrightarrow{AB} = \mathbf{a} + \frac{3}{4}(\mathbf{b} - \mathbf{a}) = \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\). Therefore, \(\overrightarrow{OX} = \mu \overrightarrow{OP} = \frac{\mu}{4}\mathbf{a} + \frac{3\mu}{4}\mathbf{b}\). Second, \(\overrightarrow{OQ} = \frac{1}{2}\mathbf{a}\), so \(\overrightarrow{BQ} = \frac{1}{2}\mathbf{a} - \mathbf{b}\). Thus, \(\overrightarrow{OX} = \overrightarrow{OB} + \lambda \overrightarrow{BQ} = \mathbf{b} + \lambda\left(\frac{1}{2}\mathbf{a} - \mathbf{b}\right) = \frac{\lambda}{2}\mathbf{a} + (1 - \lambda)\mathbf{b}\). Equating components: For \(\mathbf{a}\): \(\frac{\mu}{4} = \frac{\lambda}{2} \implies \mu = 2\lambda\). For \(\mathbf{b}\): \(\frac{3\mu}{4} = 1 - \lambda\). Substituting \(\mu = 2\lambda\) gives \(\frac{6\lambda}{4} = 1 - \lambda \implies \frac{3\lambda}{2} + \lambda = 1 \implies \frac{5\lambda}{2} = 1 \implies \lambda = \frac{2}{5}\). Then \(\mu = \frac{4}{5}\).

M1: Finding \(\overrightarrow{OP} = \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\). M1: Expressing \(\overrightarrow{OX}\) using \(\mu\). M1: Finding \(\overrightarrow{BQ} = \frac{1}{2}\mathbf{a} - \mathbf{b}\) and expressing \(\overrightarrow{OX}\) using \(\lambda\). M1: Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to set up two simultaneous equations. A1: Finding \(\lambda = 2/5\) (or 0.4). A1: Finding \(\mu = 4/5\) (or 0.8).

Let \( y = \text{f}(x) \). Then, \( y = \frac{3x + 2}{x - 4} \). Multiply both sides by \( x - 4 \) to get \( y(x - 4) = 3x + 2 \). Expand the brackets: \( xy - 4y = 3x + 2 \). Rearrange the terms to group all terms containing \( x \) on one side of the equation: \( xy - 3x = 4y + 2 \). Factor out \( x \): \( x(y - 3) = 4y + 2 \). Divide both sides by \( y - 3 \) to express \( x \) in terms of \( y \): \( x = \frac{4y + 2}{y - 3} \). Replacing \( y \) with \( x \) gives the inverse function: \( \text{f}^{-1}(x) = \frac{4x + 2}{x - 3} \). For part (b), we solve \( \text{fg}(x) = 5 \), which means \( \text{f}(\text{g}(x)) = 5 \). This implies that \( \text{g}(x) = \text{f}^{-1}(5) \). Substituting \( 5 \) into our inverse function formula from part (a): \( \text{g}(x) = \frac{4(5) + 2}{5 - 3} = \frac{22}{2} = 11 \). Since \( \text{g}(x) = x^2 - 5 \), we set up the equation \( x^2 - 5 = 11 \). Adding \( 5 \) to both sides yields \( x^2 = 16 \). Taking the square root of both sides gives the solutions \( x = 4 \) or \( x = -4 \).

Part (a): M1 for setting up the equation \( y = \frac{3x+2}{x-4} \) and attempting to make \( x \) the subject (e.g. multiplying by the denominator). M1 for correctly grouping terms in \( x \) and factoring, such as \( x(y-3) = 4y+2 \). A1 for the correct expression \( \text{f}^{-1}(x) = \frac{4x+2}{x-3} \) (must be in terms of \( x \)). Part (b): M1 for establishing the relation \( \text{g}(x) = \text{f}^{-1}(5) \) or attempting to substitute \( \text{g}(x) \) into \( \text{f}(x) \). M1 for simplifying the resulting equation to the quadratic form \( x^2 = 16 \) (or equivalent). A1 for obtaining both solutions \( x = 4 \) and \( x = -4 \) (or written as \( x = \pm 4 \)).

Time taken by the car = \(\frac{180}{x}\) hours.
Time taken by the train = \(\frac{200}{x + 15}\) hours.
Since the difference is \(40\text{ minutes} = \frac{40}{60} = \frac{2}{3}\text{ hours}\), we have:
\(\frac{180}{x} - \frac{200}{x + 15} = \frac{2}{3}\)

Multiply both sides by \(3x(x + 15)\):
\(3 \times 180(x + 15) - 3 \times 200x = 2x(x + 15)\)
\(540x + 8100 - 600x = 2x^2 + 30x\)
\(-60x + 8100 = 2x^2 + 30x\)
\(2x^2 + 90x - 8100 = 0\)

Divide by 2:
\(x^2 + 45x - 4050 = 0\) (as required)

(b) Solve the quadratic equation:
\(x = \frac{-45 \pm \sqrt{45^2 - 4(1)(-4050)}}{2}\)
\(x = \frac{-45 \pm \sqrt{2025 + 16200}}{2}\)
\(x = \frac{-45 \pm \sqrt{18225}}{2}\)
\(x = \frac{-45 \pm 135}{2}\)

Since speed must be positive, \(x = \frac{90}{2} = 45\).
So, the average speed of the car is \(45\text{ km/h}\).

M1 for writing correct expressions for the time taken by the car and train, e.g., \(\frac{180}{x}\) and \(\frac{200}{x+15}\).
M1 for writing a correct equation representing the difference in time: \(\frac{180}{x} - \frac{200}{x+15} = \frac{2}{3}\) (or equivalent).
M1 for multiplying by the common denominator to clear fractions: \(540(x+15) - 600x = 2x(x+15)\).
M1 for expanding and rearranging: \(540x + 8100 - 600x = 2x^2 + 30x\).
A1 for completing the algebra to show \(x^2 + 45x - 4050 = 0\) clearly.
M1 for attempting to solve the quadratic equation using the quadratic formula or factorisation: \((x - 45)(x + 90) = 0\).
A1 for \(x = 45\) (and rejecting \(x = -90\)).
A0.33 for final correct concluding statement: average speed of the car is \(45\text{ km/h}\).

(a) The formula for compound interest after \(t\) years is \(A = P \left(1 + \frac{r}{100}\right)^t\).
After 3 years: \(P \left(1 + \frac{r}{100}\right)^3 = 6298.56\) (Equation 1)
After 5 years: \(P \left(1 + \frac{r}{100}\right)^5 = 7346.64\) (Equation 2)

Divide Equation 2 by Equation 1:
\(\frac{P \left(1 + \frac{r}{100}\right)^5}{P \left(1 + \frac{r}{100}\right)^3} = \frac{7346.64}{6298.56}\)
\(\left(1 + \frac{r}{100}\right)^2 = 1.1664\) (as required)

Now find \(r\):
\(1 + \frac{r}{100} = \sqrt{1.1664} = 1.08\)
\(\frac{r}{100} = 0.08 \implies r = 8\)

(b) Using Equation 1:
\(P(1.08)^3 = 6298.56\)
\(P \times 1.259712 = 6298.56\)
\(P = \frac{6298.56}{1.259712} = 5000\)

(c) Total money withdrawn = \(£7346.64\).
Amount remaining after buying the laptop (spending \(15\%\)):
\(7346.64 \times (1 - 0.15) = 7346.64 \times 0.85 = 6244.644\)
Amount remaining after buying the bicycle (spending \(20\%\) of the remaining):
\(6244.644 \times (1 - 0.20) = 6244.644 \times 0.80 = 4995.7152\)
Rounding to the nearest penny: \(£4995.72\).

M1 for setting up the two compound interest equations.
M1 for dividing the equations to show \((1 + r/100)^2 = 1.1664\).
A1 for solving to find \(r = 8\).
M1 for substituting \(r = 8\) into either of the original equations.
A1 for \(P = 5000\).
M1 for calculating the amount after the laptop purchase: \(7346.64 \times 0.85\).
M1 for multiplying this result by \(0.80\).
A1.33 for \(£4995.72\) (accept \(4995.72\)).

(a) Let \(y = 2x - 3\).
\(y + 3 = 2x \implies x = \frac{y + 3}{2}\).
So, \(\mathrm{f}^{-1}(x) = \frac{x + 3}{2}\).

(b) \(\mathrm{f}(4) = 2(4) - 3 = 5\).
\(\mathrm{g}(5) = \frac{6}{5 - 1} = \frac{6}{4} = 1.5\).

(c) \(\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x)) = 2\left(\frac{6}{x - 1}\right) - 3\).
Set this equal to \(x\):
\(\frac{12}{x - 1} - 3 = x\)
\(\frac{12}{x - 1} = x + 3\)
\(12 = (x + 3)(x - 1)\)
\(12 = x^2 + 2x - 3\)
\(x^2 + 2x - 15 = 0\)
\((x + 5)(x - 3) = 0\)
Thus, \(x = 3\) or \(x = -5\).

M1 for setting \(y = 2x - 3\) and attempting to make \(x\) the subject.
A1 for \(\mathrm{f}^{-1}(x) = \frac{x + 3}{2}\).
M1 for substituting \(x = 4\) into \(\mathrm{f}(x)\) to get 5, then substituting 5 into \(\mathrm{g}(x)\).
A1 for \(1.5\).
M1 for substituting the expression for \(\mathrm{g}(x)\) into \(\mathrm{f}(x)\).
M1 for setting equal to \(x\) and forming a quadratic: \(x^2 + 2x - 15 = 0\).
A1 for solving to obtain \(x = 3\) and \(x = -5\).
A0.33 for no extra invalid values.

\(P\) is on \(OA\) such that \(OP : PA = 2 : 1 \implies \overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).
\(Q\) is on \(AB\) such that \(AQ : QB = 1 : 3 \implies \overrightarrow{AQ} = \frac{1}{4}\overrightarrow{AB} = \frac{1}{4}(\mathbf{b} - \mathbf{a})\).

(a)(i)
\(\overrightarrow{OQ} = \overrightarrow{OA} + \overrightarrow{AQ} = \mathbf{a} + \frac{1}{4}(\mathbf{b} - \mathbf{a}) = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}\)

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

(b)
Using \(\overrightarrow{OX} = \mu \overrightarrow{OQ}\):
\(\overrightarrow{OX} = \mu\left(\frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}\right) = \frac{3}{4}\mu\mathbf{a} + \frac{1}{4}\mu\mathbf{b}\)

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

Equating coefficients:
From \(\mathbf{a}\): \(\frac{3}{4}\mu = \frac{2}{3}\lambda \implies \mu = \frac{8}{9}\lambda\)
From \(\mathbf{b}\): \(\frac{1}{4}\mu = 1 - \lambda\)

Substitute \(\mu\):
\(\frac{1}{4}\left(\frac{8}{9}\lambda\right) = 1 - \lambda \implies \frac{2}{9}\lambda = 1 - \lambda \implies \frac{11}{9}\lambda = 1 \implies \lambda = \frac{9}{11}\)

Then:
\(\mu = \frac{8}{9} \times \frac{9}{11} = \frac{8}{11}\).

M1 for finding \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\) or \(\overrightarrow{AQ} = \frac{1}{4}(\mathbf{b} - \mathbf{a})\).
A1 for \(\overrightarrow{OQ} = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}\).
A1 for \(\overrightarrow{BP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\).
M1 for writing \(\overrightarrow{OX}\) in two different ways using parameters.
M1 for equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\).
M1 for solving the simultaneous equations.
A1 for \(\lambda = \frac{9}{11}\).
A0.33 for \(\mu = \frac{8}{11}\).

(a) The probability that both marbles are red is:
\(\mathrm{P}(R, R) = \frac{5}{n} \times \frac{4}{n - 1} = \frac{20}{n(n - 1)}
\).

(b)(i) We are given \(\mathrm{P}(R, R) = \frac{5}{33}\), so:
\(\frac{20}{n(n - 1)} = \frac{5}{33}\)
\(20 \times 33 = 5n(n - 1)\)
\(660 = 5n^2 - 5n\)
\(5n^2 - 5n - 660 = 0\)
Dividing both sides by 5:
\(n^2 - n - 132 = 0\) (as required).

(ii) Factorise the quadratic equation:
\((n - 12)(n + 11) = 0\)
Since \(n\) must be a positive integer, \(n = 12\).

(c) With \(n = 12\), there are 5 red marbles and \(12 - 5 = 7\) blue marbles.
\(\mathrm{P}(\text{different}) = \mathrm{P}(R, B) + \mathrm{P}(B, R)\)
\(\mathrm{P}(R, B) = \frac{5}{12} \times \frac{7}{11} = \frac{35}{132}\)
\(\mathrm{P}(B, R) = \frac{7}{12} \times \frac{5}{11} = \frac{35}{132}\)
\(\mathrm{P}(\text{different}) = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66}\).

M1 for \(\frac{5}{n} \times \frac{4}{n - 1}\).
A1 for \(\frac{20}{n(n - 1)}
\).
M1 for setting up the equation and cross-multiplying.
A1 for completing the algebra to show \(n^2 - n - 132 = 0\).
M1 for solving the quadratic equation.
A1 for choosing \(n = 12\).
M1 for calculating \(\mathrm{P}(R, B) + \mathrm{P}(B, R)\) with \(n = 12\).
A0.33 for \(\frac{35}{66}\).

The diagonal \(AC\) of the base is:
\(AC = \sqrt{12^2 + 9^2} = \sqrt{225} = 15\text{ cm}\).
Since \(O\) is the centre:
\(AO = 7.5\text{ cm}\).

(a) In right-angled triangle \(VOA\):
\(VO = \sqrt{VA^2 - AO^2} = \sqrt{15^2 - 7.5^2} = \sqrt{168.75} \approx 12.990 \approx 13.0\text{ cm}\) (to 3 s.f.).

(b) Angle between edge \(VA\) and base is \(\angle VAO\):
\(\cos(\angle VAO) = \frac{AO}{VA} = \frac{7.5}{15} = 0.5\)
\(\angle VAO = \arccos(0.5) = 60^\circ\).

(c) Let \(M\) be the midpoint of \(AB\).
In triangle \(VOM\), \(OM = \frac{1}{2} BC = 4.5\text{ cm}\).
\(\tan(\angle VMO) = \frac{VO}{OM} = \frac{\sqrt{168.75}}{4.5} \approx 2.88675\)
\(\angle VMO = \arctan(2.88675) \approx 70.89^\circ \approx 70.9^\circ\) (to 1 d.p.).

M1 for finding \(AC = 15\) or \(AO = 7.5\).
M1 for applying Pythagoras' theorem to find \(VO\).
A1 for \(VO = 13.0\) (or \(7.5\sqrt{3}\)).
M1 for identifying the correct angle \(\angle VAO\) and using trigonometry.
A1 for \(60^\circ\).
M1 for identifying the correct angle between plane and base as \(\angle VMO\) and finding \(OM = 4.5\).
M1 for using trigonometry in triangle \(VOM\).
A0.33 for \(70.9^\circ\) (accept \(70.8^\circ\) to \(71.0^\circ\)).

(a) The Venn diagram regions are:
- \(N \cap D \cap P = 12\)
- \(N \cap D\text{ only} = 22 - 12 = 10\)
- \(D \cap P\text{ only} = 18 - 12 = 6\)
- \(N \cap P\text{ only} = 20 - 12 = 8\)
- \(N\text{ only} = 3x\)
- \(D\text{ only} = x + 10\)
- \(P\text{ only} = 2x\)
- Outside the circles = 14

The sum of all these regions is 120:
\((3x) + (x + 10) + (2x) + 10 + 6 + 8 + 12 + 14 = 120\)
\(6x + 60 = 120\) (as required).

(b) Solve for \(x\):
\(6x = 60 \implies x = 10\).

(c)(i) People who use at least two services:
\(10 + 6 + 8 + 12 = 36\).

(ii) People who use Prime Video but not Netflix:
\((P\text{ only}) + (D \cap P\text{ only}) = 2x + 6 = 2(10) + 6 = 26\).

M1 for finding the 'only' intersections: \(10\), \(6\), and \(8\).
M1 for summing all the components of the Venn diagram.
A1 for completing algebra to show \(6x + 60 = 120\).
A1 for \(x = 10\).
M1 for identifying 'at least two services' regions and summing them.
A1 for \(36\).
M1 for identifying 'Prime Video but not Netflix' regions: \(2x + 6\).
A0.33 for \(26\).

(a) Volume of cylinder: \(V_{\text{cylinder}} = \pi r^2 (3r) = 3\pi r^3\).
Volume of cone: \(V_{\text{cone}} = \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3\).
Total volume:
\(V = 3\pi r^3 + \frac{2}{3}\pi r^3 = \frac{11}{3}\pi r^3\).

(b) Given \(V = 792\pi\):
\(\frac{11}{3}\pi r^3 = 792\pi \implies r^3 = 792 \times \frac{3}{11} = 216 \implies r = 6\).

(c) Total surface area:
1) Base of cylinder: \(\pi r^2 = 36\pi\).
2) Curved surface of cylinder: \(2\pi r (3r) = 216\pi\).
3) Curved surface of cone: \(\pi r l\), where \(l = \sqrt{r^2 + (2r)^2} = r\sqrt{5} = 6\sqrt{5}\).
\(A_{\text{cone}} = \pi (6)(6\sqrt{5}) = 36\sqrt{5}\pi\).
Total surface area:
\(A_{\text{total}} = 36\pi + 216\pi + 36\sqrt{5}\pi = 252\pi + 36\sqrt{5}\pi \approx 1044.57\text{ cm}^2\).
To 3 significant figures, this is \(1040\text{ cm}^2\).

M1 for finding the volume of the cylinder as \(3\pi r^3\) or the volume of the cone as \(\frac{2}{3}\pi r^3\).
A1 for adding the two volumes to show \(V = \frac{11}{3}\pi r^3\).
M1 for setting \(\frac{11}{3}\pi r^3 = 792\pi\) and solving for \(r^3\).
A1 for \(r = 6\).
M1 for calculating the cylinder base area plus cylinder curved surface area: \(252\pi\).
M1 for finding the slant height of the cone: \(l = 6\sqrt{5}\).
M1 for finding the curved surface area of the cone: \(36\sqrt{5}\pi\).
A0.33 for \(1040\) (accept \(1044 - 1045\)).

(a) The time taken for the outward journey is distance divided by speed:
\text{Time} = \frac{40}{x} \text{ hours}

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

(c) The difference in times is 40 minutes, which is:
\frac{40}{60} = \frac{2}{3} \text{ hours}

Since the return journey takes longer:
\frac{40}{x-3} - \frac{40}{x} = \frac{2}{3}

Multiply through by \(3x(x-3)\) to clear the fractions:
3 \cdot 40x - 3 \cdot 40(x-3) = 2x(x-3)
120x - 120x + 360 = 2x^2 - 6x
360 = 2x^2 - 6x

Divide both sides by 2:
180 = x^2 - 3x
x^2 - 3x - 180 = 0

(d) Solve \(x^2 - 3x - 180 = 0\):
(x - 15)(x + 12) = 0
Since speed must be positive, \(x = 15\).

The outward time is:
\frac{40}{15} = \frac{8}{3} \text{ hours} = 2 \text{ hours and } 40 \text{ minutes}.

M1: For writing \frac{40}{x} (or equivalent)
M1: For writing \frac{40}{x-3} (or equivalent)
M1: For setting up the correct difference equation \frac{40}{x-3} - \frac{40}{x} = \frac{40}{60}
A1: For correctly expanding and showing the algebraic steps to reach \(x^2 - 3x - 180 = 0\)
M1: For factorising or using the quadratic formula to solve \(x^2 - 3x - 180 = 0\)
A1: For finding \(x = 15\) (and rejecting \(x = -12\))
M1: For substituting their positive \(x\) value back to find the outward time: \frac{40}{15}
A1: For 2 hours 40 minutes

(a) \det(\mathbf{A}) = (2)(3) - (k)(-1) = 6 + k
Given \det(\mathbf{A}) = 10:
6 + k = 10 \Rightarrow k = 4

(b) With \(k = 4\), \mathbf{A} = \begin{pmatrix} 2 & 4 \\ -1 & 3 \end{pmatrix}.
2\mathbf{A} = \begin{pmatrix} 4 & 8 \\ -2 & 6 \end{pmatrix}

Calculate \mathbf{B}^2:
\mathbf{B}^2 = \begin{pmatrix} 3 & -2 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} 3 & -2 \\ 1 & 4 \end{pmatrix} = \begin{pmatrix} 3(3) + (-2)(1) & 3(-2) + (-2)(4) \\ 1(3) + 4(1) & 1(-2) + 4(4) \end{pmatrix} = \begin{pmatrix} 7 & -14 \\ 7 & 14 \end{pmatrix}

Now find \mathbf{C}:
\mathbf{C} = 2\mathbf{A} - \mathbf{B}^2 = \begin{pmatrix} 4 & 8 \\ -2 & 6 \end{pmatrix} - \begin{pmatrix} 7 & -14 \\ 7 & 14 \end{pmatrix} = \begin{pmatrix} -3 & 22 \\ -9 & -8 \end{pmatrix}

(c) The inverse matrix \mathbf{A}^{-1} is given by:
\mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})} \begin{pmatrix} 3 & -4 \\ 1 & 2 \end{pmatrix} = \frac{1}{10} \begin{pmatrix} 3 & -4 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 0.3 & -0.4 \\ 0.1 & 0.2 \end{pmatrix}

M1: For setting up the determinant equation \(2(3) - k(-1) = 10\)
A1: For \(k = 4\)
M1: For calculating \(\mathbf{B}^2\) with at least two correct elements
A1: For \(\mathbf{B}^2 = \begin{pmatrix} 7 & -14 \\ 7 & 14 \end{pmatrix}\)
M1: For performing the matrix subtraction \(2\mathbf{A} - \mathbf{B}^2\) element-wise
A1: For \(\mathbf{C} = \begin{pmatrix} -3 & 22 \\ -9 & -8 \end{pmatrix}\)
M1: For attempting to find the inverse using their \(k\): \frac{1}{10} \begin{pmatrix} 3 & -k \\ 1 & 2 \end{pmatrix}
A1: For \(\begin{pmatrix} 0.3 & -0.4 \\ 0.1 & 0.2 \end{pmatrix}\) (or fractional equivalent)

(a) To find the angle \(\angle PQR\):
The bearing of \(Q\) from \(P\) is \(060^\circ\). The back-bearing from \(Q\) to \(P\) is \(060^\circ + 180^\circ = 240^\circ\).
The bearing of \(R\) from \(Q\) is \(145^\circ\).
The angle \(\angle PQR\) is the difference between these bearings:
\angle PQR = 240^\circ - 145^\circ = 95^\circ

Using the cosine rule in triangle \(PQR\):
PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(\angle PQR)
PR^2 = 12^2 + 18^2 - 2 \cdot 12 \cdot 18 \cdot \cos(95^\circ)
PR^2 = 144 + 324 - 432 \cdot \cos(95^\circ)
PR^2 \approx 468 - 432 \cdot (-0.087156)
PR^2 \approx 468 + 37.651 = 505.651
PR = \sqrt{505.651} \approx 22.487 \text{ km} \approx 22.5 \text{ km}

(b) Using the sine rule to find \(\angle QPR\):
\frac{\sin(\angle QPR)}{QR} = \frac{\sin(\angle PQR)}{PR}
\sin(\angle QPR) = \frac{18 \cdot \sin(95^\circ)}{22.487} \approx \frac{18 \cdot 0.99619}{22.487} \approx 0.7974
\angle QPR \approx \arcsin(0.7974) \approx 52.88^\circ

The bearing of \(R\) from \(P\) is the bearing of \(Q\) from \(P\) plus \(\angle QPR\):
\text{Bearing} = 60^\circ + 52.88^\circ = 112.88^\circ \approx 113^\circ

M1: For calculating \(\angle PQR = 240^\circ - 145^\circ = 95^\circ\) (or equivalent exterior angle work)
M1: For correct application of the cosine rule: \(PR^2 = 12^2 + 18^2 - 2 \cdot 12 \cdot 18 \cdot \cos(95^\circ)\)
A1: For \(PR \approx 22.5\text{ km}\) (accept 22.4 to 22.6)
M1: For correct application of the sine rule: \(\frac{\sin(\angle QPR)}{18} = \frac{\sin(95^\circ)}{PR}\) (using their value for \(PR\))
A1: For finding \(\sin(\angle QPR) \approx 0.797\) or \(\angle QPR \approx 52.9^\circ\)
M1: For adding the initial bearing \(60^\circ\) to their \(\angle QPR\)
A1: For bearing of \(113^\circ\) (accept \(112.8^\circ\) to \(113.2^\circ\))

(a) The total number of balls is \(5 + 3 + 2 = 10\).
Since there are only 2 green balls, it is impossible to draw 3 green balls.
Therefore, the only possible ways to draw three balls of the same colour are Red-Red-Red (R-R-R) or Blue-Blue-Blue (B-B-B).

\(P(\text{R-R-R}) = \frac{5}{10} \times \frac{4}{9} \times \frac{3}{8} = \frac{60}{720} = \frac{1}{12}\)
\(P(\text{B-B-B}) = \frac{3}{10} \times \frac{2}{9} \times \frac{1}{8} = \frac{6}{720} = \frac{1}{120}\)

\(P(\text{same colour}) = \frac{60}{720} + \frac{6}{720} = \frac{66}{720} = \frac{11}{120}\) (or 0.0917)

(b) At least two red balls means either exactly two red balls or exactly three red balls.

For exactly two red balls:
The sequences are R-R-N, R-N-R, and N-R-R, where N is a non-red ball (there are 5 non-red balls).
\(P(\text{R-R-N}) = \frac{5}{10} \times \frac{4}{9} \times \frac{5}{8} = \frac{100}{720}\)
\(P(\text{R-N-R}) = \frac{5}{10} \times \frac{5}{9} \times \frac{4}{8} = \frac{100}{720}\)
\(P(\text{N-R-R}) = \frac{5}{10} \times \frac{5}{9} \times \frac{4}{8} = \frac{100}{720}\)
Sum of exactly two red balls = \frac{300}{720}

For exactly three red balls:
\(P(\text{R-R-R}) = \frac{60}{720}\)

\(P(\text{at least 2 red}) = \frac{300}{720} + \frac{60}{720} = \frac{360}{720} = \frac{1}{2}\) (or 0.5)

(c) Reduced Sample Space method:
Once the first ball is Blue, the remaining 9 balls are: 5 Red, 2 Blue, and 2 Green.
We now need the probability that the second ball chosen from this set of 9 is Green.
The pathways to a Green ball on the third draw overall (which is the second draw from the remaining 9) are Red-Green, Blue-Green, or Green-Green:
\(P(\text{R-G}) = \frac{5}{9} \times \frac{2}{8} = \frac{10}{72}\)
\(P(\text{B-G}) = \frac{2}{9} \times \frac{2}{8} = \frac{4}{72}\)
\(P(\text{G-G}) = \frac{2}{9} \times \frac{1}{8} = \frac{2}{72}\)
Sum = \frac{16}{72} = \frac{2}{9}\)

M1: For calculating \(P(\text{R-R-R}) = \frac{5}{10} \times \frac{4}{9} \times \frac{3}{8}\) or \(P(\text{B-B-B}) = \frac{3}{10} \times \frac{2}{9} \times \frac{1}{8}\)
M1: For adding the two same-colour probabilities: \(P(\text{R-R-R}) + P(\text{B-B-B})\)
A1: For \(\frac{11}{120}\) (or equivalent)
M1: For setting up the probability of exactly two reds, showing at least one of the cases (e.g. \(\frac{5}{10} \times \frac{4}{9} \times \frac{5}{8}\))
M1: For summing three permutations of exactly two reds and one three-red case: \(3 \times \frac{100}{720} + \frac{60}{720}\)
A1: For \(\frac{1}{2}\) (or 0.5)
M1: For using the reduced sample space or identifying B-R-G, B-B-G, B-G-G probabilities
A1: For \(\frac{2}{9}\) (or equivalent fraction, e.g. \(\frac{16}{72}\))