2024 Edexcel IGCSE Mathematics (Specification B) Practice Paper with Answers

Multiply the entire equation by the common denominator \(2(x - 2)(x + 2)\): \\ \(2 \times 3(x + 2) - 2 \times 2(x - 2) = (x - 2)(x + 2)\) \\ \(6(x + 2) - 4(x - 2) = x^2 - 4\) \\ \(6x + 12 - 4x + 8 = x^2 - 4\) \\ \(2x + 20 = x^2 - 4\) \\ Rearranging into standard quadratic form: \\ \(x^2 - 2x - 24 = 0\) \\ Factorising the quadratic equation: \\ \((x - 6)(x + 4) = 0\) \\ Thus, \(x = 6\) or \(x = -4\).

M1 for writing the equation with a common denominator or removing the fraction, e.g., \(6(x + 2) - 4(x - 2) = x^2 - 4\). M1 for rearranging to a 3-term quadratic equation, e.g., \(x^2 - 2x - 24 = 0\). A1 for factorising or using the formula to find roots, e.g., \((x - 6)(x + 4) = 0\). A1 for the correct solutions \(x = 6\) and \(x = -4\).

Using the determinant property \(\det(\mathbf{AB}) = \det(\mathbf{A}) \times \det(\mathbf{B})\): \\ \(\det(\mathbf{A}) = y(4) - (3)(-2) = 4y + 6\) \\ \(\det(\mathbf{B}) = 1(y) - (-1)(2) = y + 2\) \\ Thus, \((4y + 6)(y + 2) = 90\) \\ \(4y^2 + 8y + 6y + 12 = 90\) \\ \(4y^2 + 14y - 78 = 0\) \\ Divide by 2: \\ \(2y^2 + 7y - 39 = 0\) \\ Factorising this gives: \\ \((2y + 13)(y - 3) = 0\) \\ Hence, the possible values of \(y\) are \(y = 3\) and \(y = -6.5\) (or \(-\frac{13}{2}\)).

M1 for finding expressions for the determinants, e.g., \(\det(\mathbf{A}) = 4y + 6\) and \(\det(\mathbf{B}) = y + 2\) (or for performing the matrix multiplication first to get \(\mathbf{AB}\)). M1 for setting up the quadratic equation, e.g., \(4y^2 + 14y + 12 = 90\). A1 for factorising correctly, e.g., \((2y + 13)(y - 3) = 0\). A1 for both answers \(y = 3\) and \(y = -6.5\).

Let the initial value be \(V\). \\ Multiplier for Year 1: \(1.12\) \\ Multiplier for Year 2: \(1.08\) \\ Multiplier for Year 3: \(0.85\) \\ The overall multiplier after 3 years is: \\ \(1.12 \times 1.08 \times 0.85 = 1.02816\) \\ The overall percentage change is: \\ \((1.02816 - 1) \times 100 = 2.816\%\) \\ Rounding to 3 significant figures gives \(2.82\%\).

M1 for multiplying the three scale factors: \(1.12 \times 1.08 \times 0.85\) (or for calculating the value of the car after 3 years as \(\text{\pounds}25,704\)). M1 for calculating the percentage change from the original, e.g., \(\frac{25704 - 25000}{25000} \times 100\) or \((1.02816 - 1) \times 100\). A1 for \(2.82\%\) (accept 2.82).

Let \(k = \mathrm{f}^{-1}(5)\), which means \(\mathrm{f}(k) = 5\). \\ \(\frac{2k - 3}{k - 3} = 5\) \\ \(2k - 3 = 5(k - 3)\) \\ \(2k - 3 = 5k - 15\) \\ \(3k = 12 \implies k = 4\) \\ So, \(\mathrm{f}^{-1}(5) = 4\). \\ Now find \(\mathrm{g}\mathrm{f}^{-1}(5) = \mathrm{g}(4)\): \\ \(\mathrm{g}(4) = 4(4) - 3 = 13\).

M1 for setting \(\mathrm{f}(k) = 5\) and attempting to solve for \(k\), OR for attempting to find the inverse function \(\mathrm{f}^{-1}(x) = \frac{3x - 3}{x - 2}\). A1 for obtaining \(\mathrm{f}^{-1}(5) = 4\). A1 for \(13\).

The number of students who study exactly two languages is the sum of those in the three intersection regions excluding the triple intersection: \\ 1) French and German only: \(\mathrm{n}(F \cap G \cap S') = 12 - 4 = 8\) \\ 2) French and Spanish only: \(\mathrm{n}(F \cap S \cap G') = 10 - 4 = 6\) \\ 3) German and Spanish only: \(\mathrm{n}(G \cap S \cap F') = 8 - 4 = 4\) \\ Total = \(8 + 6 + 4 = 18\).

M1 for attempting to find the number of students in at least one of the double-intersection-only regions, e.g., \(12 - 4 = 8\). M1 for identifying all three two-set-only intersection values: 8, 6, and 4. A1 for 18.

1. The tangent \(TA\) is perpendicular to the radius \(OA\), so \(\angle OAT = 90^\circ\). \\ 2. In right-angled triangle \(OAT\), \(\angle AOT = 180^\circ - 90^\circ - 36^\circ = 54^\circ\). \\ 3. Since \(O\), \(C\), and \(T\) lie on a straight line, the angle at the centre is \(\angle AOC = \angle AOT = 54^\circ\). \\ 4. The angle subtended by arc \(AC\) at the circumference is half the angle subtended at the centre: \(\angle ABC = \frac{1}{2} \angle AOC = \frac{54^\circ}{2} = 27^\circ\).

M1 for identifying that \(\angle OAT = 90^\circ\). M1 for calculating \(\angle AOC = 54^\circ\) using the angle sum in triangle \(OAT\). A1 for the final answer of \(27^\circ\).

Find \(\overrightarrow{PQ} = \mathbf{q} - \mathbf{p}\). \\ Since \(PR : RQ = 3 : 2\), we have \(\overrightarrow{PR} = \frac{3}{5}(\mathbf{q} - \mathbf{p})\). \\ Then \(\overrightarrow{OR} = \overrightarrow{OP} + \overrightarrow{PR} = \mathbf{p} + \frac{3}{5}(\mathbf{q} - \mathbf{p}) = \frac{2}{5}\mathbf{p} + \frac{3}{5}\mathbf{q}\). \\ Since \(OS : SQ = 2 : 1\), we have \(\overrightarrow{OS} = \frac{2}{3}\mathbf{q}\). \\ Now find \(\overrightarrow{RS}\): \\ \(\overrightarrow{RS} = \overrightarrow{OS} - \overrightarrow{OR} = \frac{2}{3}\mathbf{q} - \left(\frac{2}{5}\mathbf{p} + \frac{3}{5}\mathbf{q}\right) = -\frac{2}{5}\mathbf{p} + \left(\frac{2}{3} - \frac{3}{5}\right)\mathbf{q} = -\frac{2}{5}\mathbf{p} + \frac{1}{15}\mathbf{q}\).

M1 for finding \(\overrightarrow{PQ} = \mathbf{q} - \mathbf{p}\) and writing an expression for \(\overrightarrow{PR}\) or \(\overrightarrow{OR}\). M1 for finding both \(\overrightarrow{OR} = \frac{2}{5}\mathbf{p} + \frac{3}{5}\mathbf{q}\) and \(\overrightarrow{OS} = \frac{2}{3}\mathbf{q}\). A1 for \(-\frac{2}{5}\mathbf{p} + \frac{1}{15}\mathbf{q}\) (or \(\frac{1}{15}(\mathbf{q} - 6\mathbf{p})\)).

The total number of sweets in the bag is \(5 + n\). \\ Probability of selecting two red sweets is: \\ \(\frac{5}{5+n} \times \frac{4}{4+n} = \frac{20}{(5+n)(4+n)}\). \\ Setting this equal to \(\frac{2}{9}\): \\ \(\frac{20}{(5+n)(4+n)} = \frac{2}{9}\) \\ Cross-multiplying: \\ \(180 = 2(5+n)(4+n)\) \\ \(90 = n^2 + 9n + 20\) \\ \(n^2 + 9n - 70 = 0\) \\ Factorising the quadratic equation: \\ \((n + 14)(n - 5) = 0\) \\ Since \(n\) must be positive, \(n = 5\).

M1 for writing an expression for the probability of selecting two red sweets: \(\frac{5}{5+n} \times \frac{4}{4+n}\). M1 for forming and expanding/rearranging the equation into standard quadratic form: \(n^2 + 9n - 70 = 0\). A1 for factorising and solving to obtain \(n = 5\) (with negative solution discarded).

Using the properties of determinants, \(\det(\mathbf{AB}) = \det(\mathbf{A}) \times \det(\mathbf{B})\).

First, calculate the determinant of \(\mathbf{A}\):
\(\det(\mathbf{A}) = (2)(3) - (k)(-1) = 6 + k\)

Next, calculate the determinant of \(\mathbf{B}\):
\(\det(\mathbf{B}) = (1)(-1) - (4)(2) = -1 - 8 = -9\)

Using the determinant of the product:
\(\det(\mathbf{AB}) = (6 + k)(-9) = -63\)

Divide both sides by \(-9\):
\(6 + k = 7\)
\(k = 1\)

M1: for evaluating the determinant of \(\mathbf{A}\) as \(6+k\) or for correctly multiplying the matrices to find \(\mathbf{AB} = \begin{pmatrix} 2+2k & 8-k \\ 5 & -7 \end{pmatrix}\).
M1: for setting up the equation \(-9(6+k) = -63\) or \(-7(2+2k) - 5(8-k) = -63\).
A1: for \(k = 1\).

From the first equation, make \(y\) the subject:
\(3x - 13 = 2y \implies y = \frac{3x - 13}{2}\)

Substitute this expression for \(y\) into the second equation:
\(x^2 + x\left(\frac{3x - 13}{2}\right) = 6\)

Multiply the entire equation by 2 to clear the fraction:
\(2x^2 + x(3x - 13) = 12\)
\(2x^2 + 3x^2 - 13x - 12 = 0\)
\(5x^2 - 13x - 12 = 0\)

Factorise the quadratic equation:
\((5x + 4)(x - 3) = 0\)

This gives \(x = 3\) or \(x = -0.8\).

Substitute these values back to find \(y\):
If \(x = 3\), \(y = \frac{3(3) - 13}{2} = -2\).
If \(x = -0.8\), \(y = \frac{3(-0.8) - 13}{2} = -7.7\).

M1: for expressing \(y\) in terms of \(x\) (or vice versa) and substituting into the second equation.
M1: for forming a correct three-term quadratic equation, e.g., \(5x^2 - 13x - 12 = 0\).
M1: (dep on previous M1) for solving their quadratic equation to find two values of \(x\) (or \(y\)).
A1: for both correct pairs of solutions: \(x=3, y=-2\) and \(x=-0.8, y=-7.7\).

Let \(k = 1 + \frac{r}{100}\).
Using the compound interest formula:
At 3 years: \(P k^3 = 2000\)
At 6 years: \(P k^6 = 2662\)

Divide the second equation by the first:
\(\frac{P k^6}{P k^3} = \frac{2662}{2000}\)
\(k^3 = 1.331\)

Take the cube root of both sides:
\(k = \sqrt[3]{1.331} = 1.1\)

Since \(k = 1 + \frac{r}{100}\):
\(1 + \frac{r}{100} = 1.1\)
\(\frac{r}{100} = 0.1\)
\(r = 10\)

M1: for setting up the two compound interest equations: \(P(1 + r/100)^3 = 2000\) and \(P(1 + r/100)^6 = 2662\).
M1: for dividing the two equations to eliminate \(P\), yielding \((1 + r/100)^3 = 1.331\) or equivalent.
M1: for solving for \((1 + r/100)\) by taking the cube root.
A1: for \(r = 10\).

The position vector of \(P\) can be found using the vector path:
\(\vec{OP} = \vec{OA} + \frac{2}{3}\vec{AB}\)

First, calculate \(\vec{AB}\):
\(\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 4 \\ -3 \end{pmatrix} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}\)

Now, calculate \(\vec{OP}\):
\(\vec{OP} = \begin{pmatrix} 4 \\ -3 \end{pmatrix} + \frac{2}{3}\begin{pmatrix} -6 \\ 8 \end{pmatrix} = \begin{pmatrix} 4 \\ -3 \end{pmatrix} + \begin{pmatrix} -4 \\ \frac{16}{3} \end{pmatrix} = \begin{pmatrix} 0 \\ \frac{7}{3} \end{pmatrix}\)

M1: for finding \(\vec{AB} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}\) or stating a correct vector path such as \(\vec{OP} = \frac{1}{3}\vec{OA} + \frac{2}{3}\vec{OB}\).
M1: for substituting the given vector coordinates into a correct path equation.
A1: for \(\begin{pmatrix} 0 \\ \frac{7}{3} \end{pmatrix}\) or equivalent fractional/decimal components (e.g., \(y \approx 2.33\)).

Let \(x\) be the number of students who study both Physics and Chemistry.
The number of students who study Physics only is \(35 - x\).
The number of students who study Chemistry only is \(28 - x\).

The total number of students in the group is 60. Therefore:
\((35 - x) + x + (28 - x) + 12 = 60\)

Simplify the equation:
\(75 - x = 60\)
\(x = 15\)

So, 15 students study both Physics and Chemistry.

M1: for expressing the total number of students in the union as \(60 - 12 = 48\) or setting up the algebraic equation \((35 - x) + x + (28 - x) + 12 = 60\).
M1: for simplifying the linear equation to the form \(75 - x = 60\) or \(63 - x = 48\).
A1: for 15.

Set \(y = \mathrm{f}(x)\):
\(y = \frac{2x + 5}{x - 3}\)

Multiply both sides by \(x - 3\):
\(y(x - 3) = 2x + 5\)
\(xy - 3y = 2x + 5\)

Rearrange to group all terms with \(x\) on one side:
\(xy - 2x = 3y + 5\)

Factorise \(x\) out of the left-hand side:
\(x(y - 2) = 3y + 5\)

Divide by \(y - 2\):
\(x = \frac{3y + 5}{y - 2}\)

Therefore, the inverse function is:
\(\mathrm{f}^{-1}(x) = \frac{3x + 5}{x - 2}\)

M1: for setting \(y = \frac{2x + 5}{x - 3}\) (or swap \(x\) and \(y\) first) and multiplying both sides by the denominator.
M1: for isolating the terms containing \(x\) on one side and factorising to get \(x(y - 2) = 3y + 5\) (or equivalent).
A1: for \(\mathrm{f}^{-1}(x) = \frac{3x + 5}{x - 2}\) (accept \(y = \frac{3x + 5}{x - 2}\)).

The interior angle of the regular polygon is \(156^\circ\).
Therefore, the exterior angle is:
\(180^\circ - 156^\circ = 24^\circ\)

Since the sum of the exterior angles of any polygon is \(360^\circ\), the number of sides \(n\) is:
\(n = \frac{360^\circ}{24^\circ} = 15\)

M1: for calculating the exterior angle as \(180^\circ - 156^\circ = 24^\circ\) or setting up the equation \(\frac{(n - 2) \times 180}{n} = 156\).
M1: for dividing \(360\) by their exterior angle or solving the equation for \(n\).
A1: for 15.

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

Substitute the given values into the formula:
\(BC^2 = 7^2 + 9^2 - 2(7)(9)\cos(60^\circ)\)
\(BC^2 = 49 + 81 - 126 \times 0.5\)
\(BC^2 = 130 - 63 = 67\)

Take the square root of both sides:
\(BC = \sqrt{67} \approx 8.18535 \text{ cm}\)

To 3 significant figures, the length of \(BC\) is \(8.19 \text{ cm}\).

M1: for a correct substitution of given values into the Cosine Rule formula, i.e., \(BC^2 = 7^2 + 9^2 - 2(7)(9)\cos(60^\circ)\).
M1: for evaluating \(BC^2 = 67\).
A1: for 8.19 (accept values in the range 8.18 to 8.19).

We multiply the matrix \(\mathbf{A}\) by \(\mathbf{B}\): \(\mathbf{A}\mathbf{B} = \begin{pmatrix} 2(4) + x(2) \\ -1(4) + 3(2) \end{pmatrix} = \begin{pmatrix} 8 + 2x \\ 2 \end{pmatrix}\). We are given that \(\mathbf{A}\mathbf{B} = \begin{pmatrix} 14 \\ y \end{pmatrix}\). Equating the corresponding elements of the matrices: 1) \(8 + 2x = 14 \implies 2x = 6 \implies x = 3\). 2) \(y = 2\).

M1: for attempting matrix multiplication to get at least one correct expression: \(8 + 2x\) or \(-4 + 6\). M1: for setting up the equation \(8 + 2x = 14\) and solving for \(x\). A1: for \(x = 3\). A1: for \(y = 2\).

First, we find \(n(A \cup B)\) using the complement: \(n(A \cup B) = n(\mathcal{U}) - n(A \cup B)' = 60 - 10 = 50\). Next, we use the principle of inclusion-exclusion: \(n(A \cup B) = n(A) + n(B) - n(A \cap B)\). Substituting the known values: \(50 = 32 + 28 - n(A \cap B)\) which simplifies to \(50 = 60 - n(A \cap B)\). Therefore, \(n(A \cap B) = 60 - 50 = 10\).

M1: for calculating \(n(A \cup B) = 60 - 10 = 50\). M1: for substituting values into the inclusion-exclusion formula \(50 = 32 + 28 - n(A \cap B)\). A1: for the correct final answer of 10.

Multiply all terms by the common denominator \(x(x-1)\) to clear the fractions: \(6x - 4(x-1) = x(x-1)\). Expand both sides: \(6x - 4x + 4 = x^2 - x\). Simplify: \(2x + 4 = x^2 - x\). Rearrange into a standard quadratic equation: \(x^2 - 3x - 4 = 0\). Factorise the quadratic expression: \((x-4)(x+1) = 0\). This gives the solutions: \(x = 4\) or \(x = -1\).

M1: for multiplying by the common denominator to obtain a linear/quadratic equation without fractions, e.g., \(6x - 4(x-1) = x(x-1)\). M1: for expanding and rearranging into standard quadratic form \(x^2 - 3x - 4 = 0\). M1: for factorising or solving the quadratic. A1: for both correct solutions \(x = 4\) and \(x = -1\).

Using the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). Substitute the given values: \(AC^2 = 8^2 + 12^2 - 2(8)(12)\cos(150^\circ)\). Evaluate the terms: \(AC^2 = 64 + 144 - 192(-\cos(30^\circ))\) since \(\cos(150^\circ) = -\frac{\sqrt{3}}{2}\). This gives \(AC^2 = 208 - 192(-0.8660) = 208 + 166.28 = 374.28\). Taking the square root: \(AC = \sqrt{374.28} \approx 19.346\text{ cm}\). To 3 significant figures, the length of \(AC\) is \(19.3\text{ cm}\).

M1: for substituting correctly into the Cosine Rule formula: \(AC^2 = 8^2 + 12^2 - 2(8)(12)\cos(150^\circ)\). M1: for evaluating \(AC^2\) to a value around \(374.3\). A1: for the final correct answer of 19.3 (accept 19.3 to 19.4).

The number of years from the start of 2020 to the start of 2023 is \(3\) years. Since the value increases by \(8\%\) compounded annually, we use the compound interest formula: \(V = P(1 + r)^n\). Here, \(P = 25000\), \(r = 0.08\), and \(n = 3\). \(V = 25000 \times (1.08)^3 = 25000 \times 1.259712 = 31492.8\). Rounding to the nearest dollar, we get \(\$31,493\).

M1: for recognizing the compound interest structure, e.g. multiplying \(25000\) by \(1.08\) or equivalent. M1: for raising the multiplier to the power of 3, e.g. \(25000 \times (1.08)^3\). A1: for calculating \(31492.8\). A1: for rounding correctly to the nearest dollar to get 31493.

First, calculate the vector \(3\mathbf{a} + 2\mathbf{b}\): \(3\mathbf{a} + 2\mathbf{b} = 3\begin{pmatrix} 3 \\ -4 \end{pmatrix} + 2\begin{pmatrix} -2 \\ 5 \end{pmatrix} = \begin{pmatrix} 9 \\ -12 \end{pmatrix} + \begin{pmatrix} -4 \\ 10 \end{pmatrix} = \begin{pmatrix} 9 - 4 \\ -12 + 10 \end{pmatrix} = \begin{pmatrix} 5 \\ -2
\end{pmatrix}\). Next, find the magnitude: \(|3\mathbf{a} + 2\mathbf{b}| = \sqrt{5^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29}\). Evaluating this: \(\sqrt{29} \approx 5.385\). Rounding to 1 decimal place gives \(5.4\).

M1: for calculating the combined vector \(3\mathbf{a} + 2\mathbf{b}\) correctly as \(\begin{pmatrix} 5 \\ -2 \end{pmatrix}\). M1: for applying Pythagoras' theorem to find the magnitude, e.g., \(\sqrt{5^2 + (-2)^2}\). A1: for the correct final answer of 5.4.

Let \(\text{f}^{-1}(4) = x\). This is equivalent to finding the value of \(x\) such that \(\text{f}(x) = 4\). So, we solve the equation: \(\frac{3x + 1}{x - 2} = 4\). Multiply both sides by \(x - 2\): \(3x + 1 = 4(x - 2)\). Expand the right-hand side: \(3x + 1 = 4x - 8\). Rearrange the equation to solve for \(x\): \(1 + 8 = 4x - 3x \implies x = 9\). Therefore, \(\text{f}^{-1}(4) = 9\).

M1: for setting up the equation \(\frac{3x + 1}{x - 2} = 4\) (or attempting to find the expression for \(\text{f}^{-1}(x)\)). M1: for isolating \(x\), e.g., \(3x + 1 = 4x - 8\). A1: for the correct value of 9.

The total number of marbles is \(5 + 7 = 12\). There are two mutually exclusive ways to draw two marbles of the same color: both are red (RR) or both are blue (BB). 1) Probability of RR: \(P(RR) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}\). 2) Probability of BB: \(P(BB) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132}\). The total probability of getting both of the same color is: \(P(RR) + P(BB) = \frac{20}{132} + \frac{42}{132} = \frac{62}{132}\). Simplifying the fraction by dividing both numerator and denominator by 2 gives \(\frac{31}{66}\).

M1: for a correct probability product for either RR or BB, e.g., \(\frac{5}{12} \times \frac{4}{11}\) or \(\frac{7}{12} \times \frac{6}{11}\). M1: for adding the two correct products: \(\frac{20}{132} + \frac{42}{132}\). A1: for simplifying the fraction to \(\frac{31}{66}\).

Since \(\mathbf{M}\) is a singular matrix, its determinant is equal to zero: \(\det(\mathbf{M}) = (k+3)(k) - (2)(5) = 0\). Expanding the expression gives \(k^2 + 3k - 10 = 0\). Factorising the quadratic equation gives \((k+5)(k-2) = 0\). Therefore, the possible values of \(k\) are \(k = -5\) or \(k = 2\).

M1: Set up the determinant expression and equate to zero: \((k+3)k - 2 \times 5 = 0\). M1: Expand to form a correct 3-term quadratic equation: \(k^2 + 3k - 10 = 0\). A1: Solve the quadratic equation by factorisation or formula to get \((k+5)(k-2) = 0\). A1: Correct values: \(k = 2\) and \(k = -5\).

First, find the vector \(\overrightarrow{OC}\): Since \(OC : CB = 1 : 3\), we have \(\overrightarrow{OC} = \frac{1}{4}\overrightarrow{OB} = \frac{1}{4}\mathbf{b}\). Next, find \(\overrightarrow{OD}\): Since \(AD : DB = 2 : 1\), we have \(\overrightarrow{AD} = \frac{2}{3}\overrightarrow{AB} = \frac{2}{3}(\mathbf{b} - \mathbf{a})\). Now calculate \(\overrightarrow{OD}\): \(\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{AD} = \mathbf{a} + \frac{2}{3}(\mathbf{b} - \mathbf{a}) = \frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{b}\). Finally, find \(\overrightarrow{CD}\): \(\overrightarrow{CD} = -\overrightarrow{OC} + \overrightarrow{OD} = -\frac{1}{4}\mathbf{b} + \frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{b} = \frac{1}{3}\mathbf{a} + \frac{5}{12}\mathbf{b}\).

M1: Find a correct expression for either \(\overrightarrow{OC} = \frac{1}{4}\mathbf{b}\) or \(\overrightarrow{AD} = \frac{2}{3}(\mathbf{b} - \mathbf{a})\). M1: Find a correct expression for \(\overrightarrow{OD} = \frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{b}\) or use a correct vector path for \(\overrightarrow{CD}\). A1: Formulate the vector sum for \(\overrightarrow{CD}\) correctly. A1: Simplify completely to get \(\frac{1}{3}\mathbf{a} + \frac{5}{12}\mathbf{b}\).

(a) \(\mathbf{A}^2 = \begin{pmatrix} 2 & k \\ -1 & 3 \end{pmatrix} \begin{pmatrix} 2 & k \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 2(2) + k(-1) & 2(k) + k(3) \\ -1(2) + 3(-1) & -1(k) + 3(3) \end{pmatrix} = \begin{pmatrix} 4 - k & 5k \\ -5 & 9 - k \end{pmatrix}\).

(b) \(\det(\mathbf{B}) = (1)(4) - (-2)(3) = 4 + 6 = 10\).
\(\det(\mathbf{A}) = (2)(3) - (k)(-1) = 6 + k\).
\(\det(\mathbf{A}\mathbf{B}) = \det(\mathbf{A}) \det(\mathbf{B}) = 10(6 + k)\).
Given that \(\det(\mathbf{A}\mathbf{B}) = 30\), we have:
\(10(6 + k) = 30 \implies 6 + k = 3 \implies k = -3\).

(c) \(\mathbf{B}^{-1} = \frac{1}{\det(\mathbf{B})} \begin{pmatrix} 4 & 2 \\ -3 & 1 \end{pmatrix} = \frac{1}{10} \begin{pmatrix} 4 & 2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 0.4 & 0.2 \\ -0.3 & 0.1 \end{pmatrix}\).

(a) M1 for attempting to multiply \(\mathbf{A}\) by itself, with at least two correct elements.
A1 for a 2x2 matrix with diagonal elements \(4 - k\) and \(9 - k\).
A1 for non-diagonal elements \(5k\) and \(-5\) correct.

(b) M1 for finding \(\det(\mathbf{B}) = 10\) or setting up \(\det(\mathbf{A}\mathbf{B}) = \det(\mathbf{A})\det(\mathbf{B})\).
M1 for setting up the equation \(10(6 + k) = 30\) (or equivalent via direct multiplication of \(\mathbf{A}\mathbf{B}\)).
A1 for \(k = -3\).

(c) M1 for a correct attempt at the adjugate matrix or dividing by their determinant.
A1 for \(\begin{pmatrix} 0.4 & 0.2 \\ -0.3 & 0.1 \end{pmatrix}\) or equivalent fraction form.

(a) Let \(y = 3x - 5\).
Then \(y + 5 = 3x \implies x = \frac{y + 5}{3}\).
Thus, \(g^{-1}(x) = \frac{x + 5}{3}\).

(b) \(g(2) = 3(2) - 5 = 1\).
Set \(f(x) = 1\):
\(\frac{2x + 3}{x - 1} = 1 \implies 2x + 3 = x - 1 \implies x = -4\).

(c) \(fg(x) = f(g(x)) = f(3x - 5) = \frac{2(3x - 5) + 3}{(3x - 5) - 1} = \frac{6x - 10 + 3}{3x - 6} = \frac{6x - 7}{3x - 6}\).

(a) M1 for setting \(y = 3x - 5\) and attempting to make \(x\) the subject.
A1 for \(g^{-1}(x) = \frac{x + 5}{3}\) (or equivalent).

(b) M1 for finding \(g(2) = 1\).
M1 for setting \(\frac{2x + 3}{x - 1} = 1\) and attempting to solve for \(x\).
A1 for \(x = -4\).

(c) M1 for substituting \(g(x)\) into \(f(x)\).
M1 for expanding and simplifying numerator and denominator.
A1 for \(\frac{6x - 7}{3x - 6}\).

(a) The total number of counters is \(n + 8\).
The probability of selecting two blue counters without replacement is:
\(\text{P(Blue, Blue)} = \frac{8}{n + 8} \times \frac{7}{n + 7} = \frac{56}{(n + 8)(n + 7)}\).
We are given this is equal to \(\frac{14}{39}\):
\(\frac{56}{n^2 + 15n + 56} = \frac{14}{39}\).
Divide both sides of the numerator by 14:
\(\frac{4}{n^2 + 15n + 56} = \frac{1}{39}\)
\(4 \times 39 = n^2 + 15n + 56 \implies 156 = n^2 + 15n + 56\)
\(n^2 + 15n - 100 = 0\) (as required).

(b) Factorising the quadratic equation:
\((n + 20)(n - 5) = 0\).
Since the number of counters \(n\) must be positive, \(n = 5\).

(c) With \(n = 5\), there are 5 red counters and 8 blue counters (total 13 counters).
The probability of choosing different colours is:
\(\text{P(Red, Blue)} + \text{P(Blue, Red)} = \left(\frac{5}{13} \times \frac{8}{12}\right) + \left(\frac{8}{13} \times \frac{5}{12}\right) = \frac{40}{156} + \frac{40}{156} = \frac{80}{156} = \frac{20}{39}\).

(a) M1 for expressing \(\text{P(Blue, Blue)} = \frac{8}{n+8} \times \frac{7}{n+7}\).
M1 for equating their product to \(\frac{14}{39}\).
M1 for expanding \((n+8)(n+7)\) and attempting to clear fractions.
A1 for showing the steps leading to \(n^2 + 15n - 100 = 0\) with no errors.

(b) M1 for attempting to factorise or solve the quadratic equation.
A1 for \(n = 5\) (must reject \(n = -20\)).

(c) M1 for setting up the sum of probabilities for different colours: \(\frac{5}{13} \times \frac{8}{12} + \frac{8}{13} \times \frac{5}{12}\) (or \(1 - \text{P(Blue, Blue)} - \text{P(Red, Red)}\)).
A1 for \(\frac{20}{39}\) (or equivalent fraction/decimal).

(a) The total surface area of a solid cylinder is:
\(A = 2\pi r^2 + 2\pi r h = 120\pi\).
Divide the entire equation by \(2\pi\):
\(r^2 + r h = 60\).
Subtract \(r^2\) from both sides:
\(r h = 60 - r^2\).
Divide by \(r\):
\(h = \frac{60 - r^2}{r}\) (as required).

(b) The volume of a cylinder is:
\(V = \pi r^2 h\).
Substitute the expression for \(h\) from part (a):
\(V = \pi r^2 \left(\frac{60 - r^2}{r}\right) = \pi r (60 - r^2) = \pi(60r - r^3)\) (as required).

(c) To find the maximum volume, differentiate \(V\) with respect to \(r\):
\(\frac{dV}{dr} = \pi(60 - 3r^2)\).
Set \(\frac{dV}{dr} = 0\):
\(\pi(60 - 3r^2) = 0 \implies 60 - 3r^2 = 0 \implies 3r^2 = 60 \implies r^2 = 20\).
Since radius must be positive:
\(r = \sqrt{20} = 2\sqrt{5}\).

(a) M1 for using the formula for the total surface area of a solid cylinder: \(2\pi r^2 + 2\pi r h = 120\pi\).
M1 for isolating the term with \(h\).
A1 for obtaining \(h = \frac{60 - r^2}{r}\) convincingly.

(b) M1 for substituting their expression for \(h\) into \(V = \pi r^2 h\).
A1 for showing \(V = \pi(60r - r^3)\).

(c) M1 for differentiating to obtain \(\frac{dV}{dr} = \pi(60 - 3r^2)\).
M1 for setting their derivative to 0 and solving for \(r^2\).
A1 for \(r = 2\sqrt{5}\).

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

(ii) 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) Expressing \(\overrightarrow{OX}\) using both paths:
Path 1: \(\overrightarrow{OX} = \mu \overrightarrow{OQ} = \frac{1}{2}\mu\mathbf{a} + \frac{1}{2}\mu\mathbf{b}\).
Path 2: \(\overrightarrow{OX} = \overrightarrow{OB} + \overrightarrow{BX} = \mathbf{b} + \lambda \overrightarrow{BP} = \mathbf{b} + \lambda\left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\lambda\mathbf{a} + (1 - \lambda)\mathbf{b}\).
Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):
1) \(\frac{1}{2}\mu = \frac{2}{3}\lambda \implies \mu = \frac{4}{3}\lambda\)
2) \(\frac{1}{2}\mu = 1 - \lambda\)
Substitute (1) into (2):
\(\frac{2}{3}\lambda = 1 - \lambda \implies \frac{5}{3}\lambda = 1 \implies \lambda = \frac{3}{5}\).
Then \(\mu = \frac{4}{3}\left(\frac{3}{5}\right) = \frac{4}{5}\).

(c) Since \(\overrightarrow{OX} = \frac{4}{5}\overrightarrow{OQ}\), the point \(X\) divides the line \(OQ\) in the ratio \(4 : 1\).
Thus, \(OX : XQ = 4 : 1\).

(a) B1 for \(\overrightarrow{OQ} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\) (or equivalent).
M1 for finding \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).
A1 for \(\overrightarrow{BP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\) (or equivalent).

(b) M1 for expressing \(\overrightarrow{OX}\) in terms of \(\mu\) and in terms of \(\lambda\).
M1 for equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to form two simultaneous equations.
M1 for solving the simultaneous equations for \(\lambda\) or \(\mu\).
A1 for \(\mu = \frac{4}{5}\) and \(\lambda = \frac{3}{5}\).

(c) B1 for \(4 : 1\) (or equivalent).

(a) For the equation to have real roots, the discriminant \(D\) must be greater than or equal to 0:
\(D = b^2 - 4ac \ge 0\).
Here, \(a = k - 2\), \(b = 2k\), and \(c = k + 3\).
\(D = (2k)^2 - 4(k - 2)(k + 3) \ge 0\)
\(4k^2 - 4(k^2 + k - 6) \ge 0\)
\(4k^2 - 4k^2 - 4k + 24 \ge 0\)
\(-4k + 24 \ge 0 \implies 4k \le 24 \implies k \le 6\) (as required).

(b) If \(k = 2\), the coefficient of \(x^2\) becomes \(2 - 2 = 0\). A quadratic equation must have a non-zero coefficient for \(x^2\) (i.e., \(a \neq 0\)), otherwise it becomes a linear equation.

(c) Substitute \(k = 4\) into the original equation:
\((4-2)x^2 + 2(4)x + (4+3) = 0 \implies 2x^2 + 8x + 7 = 0\).
Using the quadratic formula:
\(x = \frac{-8 \pm \sqrt{8^2 - 4(2)(7)}}{2(2)} = \frac{-8 \pm \sqrt{64 - 56}}{4} = \frac{-8 \pm \sqrt{8}}{4}\).
Since \(\sqrt{8} = 2\sqrt{2}\):
\(x = \frac{-8 \pm 2\sqrt{2}}{4} = \frac{-4 \pm \sqrt{2}}{2}\).

(a) M1 for identifying the condition for real roots as \(b^2 - 4ac \ge 0\).
M1 for substituting \(a, b, c\) into the discriminant expression.
M1 for expanding and simplifying the inequality to \(-4k + 24 \ge 0\).
A1 for solving the inequality to show \(k \le 6\) with no errors.

(b) B1 for mentioning that the \(x^2\) term vanishes or the equation becomes linear.

(c) M1 for substituting \(k = 4\) to get \(2x^2 + 8x + 7 = 0\).
M1 for applying the quadratic formula correctly.
A1 for \(\frac{-4 \pm \sqrt{2}}{2}\) or equivalent fully simplified form.

(a) The area of a triangle is given by \(\frac{1}{2} a c \sin B\):
\(\text{Area} = \frac{1}{2} (AB)(BC) \sin(60^\circ) = 10\sqrt{3}\).
\(\frac{1}{2} (x + 3)(2x - 1) \left(\frac{\sqrt{3}}{2}\right) = 10\sqrt{3}\).
Multiply both sides by \(\frac{4}{\sqrt{3}}\):
\((x + 3)(2x - 1) = 40\).
Expand the left-hand side:
\(2x^2 - x + 6x - 3 = 40\)
\(2x^2 + 5x - 3 = 40 \implies 2x^2 + 5x - 43 = 0\) (as required).

(b) Using the quadratic formula for \(2x^2 + 5x - 43 = 0\):
\(x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-43)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 344}}{4} = \frac{-5 \pm \sqrt{369}}{4}\).
\(x = \frac{-5 + 19.209}{4} \approx 3.552\) or \(x = \frac{-5 - 19.209}{4} < 0\).
Since length must be positive, \(x \approx 3.55\) (to 3 s.f.).

(c) Using the Cosine Rule to find \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(60^\circ)\).
Using the more precise value of \(x = 3.552\):
\(AB = 3.552 + 3 = 6.552\) cm.
\(BC = 2(3.552) - 1 = 6.104\) cm.
\(AC^2 = 6.552^2 + 6.104^2 - 2(6.552)(6.104)(0.5)\)
\(AC^2 = 42.929 + 37.259 - 39.993 = 40.195\).
\(AC = \sqrt{40.195} \approx 6.34\) cm (to 3 s.f.).

(a) M1 for using \(\frac{1}{2} a c \sin B\) with given values.
M1 for substituting \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\) and setting up the equation.
M1 for expanding \((x+3)(2x-1)\).
A1 for showing \(2x^2 + 5x - 43 = 0\) with clear intermediate steps.

(b) M1 for attempting to solve the quadratic equation using the formula.
A1 for \(x \approx 3.55\) (accept 3.55 or 3.552, must reject the negative root).

(c) M1 for writing down a correct cosine rule formula to find \(AC^2\).
A1 for \(AC \approx 6.34\) (accept range 6.33 to 6.35).

(a) We have:
\(\text{Year 10} : \text{Year 11} = 5 : 4\)
\(\text{Year 11} : \text{Year 12} = 3 : 2\)
To combine these, find a common multiple for Year 11, which is 12:
\(\text{Year 10} : \text{Year 11} = 15 : 12\)
\(\text{Year 11} : \text{Year 12} = 12 : 8\)
Thus, the combined ratio is \(15 : 12 : 8\).

(b) There are 360 students in Year 11, which corresponds to 12 parts in the ratio.
Value of 1 part = \(\frac{360}{12} = 30\) students.
Total parts = \(15 + 12 + 8 = 35\) parts.
Total number of students = \(35 \times 30 = 1050\).

(c) The number of students in Year 12 is 8 parts:
\(\text{Year 12 students} = 8 \times 30 = 240\).
Number of Year 12 students who study Mathematics:
\(240 \times 25\% = 240 \times 0.25 = 60\).
Since 60% of these students are boys, the remaining 40% are girls:
\(\text{Number of girls} = 60 \times 40\% = 60 \times 0.4 = 24\).

(a) M1 for finding a common ratio for Year 11 (e.g. 12) or equivalent scaling.
A1 for \(15 : 12 : 8\) (or equivalent simplified form).

(b) M1 for dividing 360 by 12 (or equivalent method to find the value of one part).
M1 for finding the total number of parts (35) and multiplying by their unit value.
A1 for \(1050\).

(c) M1 for finding the total number of students in Year 12 as 240.
M1 for finding 25% of 240 (60) and then attempting to find 40% of that number.
A1 for \(24\).

(a) \(\mathbf{AB} = \begin{pmatrix} x & 2 \\ -1 & y \end{pmatrix} \begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 3x + 2 & 4x + 4 \\ -3 + y & -4 + 2y \end{pmatrix}\). Equating this to the given product: \(3x + 2 = 14 \implies 3x = 12 \implies x = 4\). Also, \(-3 + y = -1 \implies y = 2\). (b) The determinant of \(\mathbf{B}\) is \(\det(\mathbf{B}) = 3(2) - 4(1) = 2\). The inverse is \(\mathbf{B}^{-1} = \frac{1}{2} \begin{pmatrix} 2 & -4 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ -0.5 & 1.5 \end{pmatrix}\). (c) From \(\mathbf{CB} = \begin{pmatrix} 5 & 8 \\ 2 & 2 \end{pmatrix}\), multiplying by \(\mathbf{B}^{-1}\) on the right gives \(\mathbf{C} = \begin{pmatrix} 5 & 8 \\ 2 & 2 \end{pmatrix} \mathbf{B}^{-1} = \begin{pmatrix} 5 & 8 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ -0.5 & 1.5 \end{pmatrix} = \begin{pmatrix} 5(1) + 8(-0.5) & 5(-2) + 8(1.5) \\ 2(1) + 2(-0.5) & 2(-2) + 2(1.5) \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 1 & -1 \end{pmatrix}\).

(a) M1 for attempting to multiply the two matrices to find at least two correct elements in terms of \(x\) and \(y\). A1 for \(x = 4\). A1 for \(y = 2\). (b) M1 for finding the determinant of \(\mathbf{B}\) (\(2\)) or writing the adjoint matrix. A1 for the correct inverse matrix. (c) M1 for attempting to multiply the matrix by their \(\mathbf{B}^{-1}\) in the correct order. A1 for the correct matrix \(\mathbf{C}\).

(a) The total number of parts in mix A is \(2 + 3 + 5 = 10\). The mass of cement in \(750\) kg is \(750 \times \frac{2}{10} = 150\) kg. (b) For \(600\) kg of mix A: mass of cement = \(600 \times \frac{2}{10} = 120\) kg, mass of sand = \(600 \times \frac{3}{10} = 180\) kg, mass of gravel = \(600 \times \frac{5}{10} = 300\) kg. For \(1200\) kg of mix B: total parts = \(12\). Mass of cement = \(1200 \times \frac{3}{12} = 300\) kg, mass of sand = \(1200 \times \frac{4}{12} = 400\) kg, mass of gravel = \(1200 \times \frac{5}{12} = 500\) kg. In mix C: total cement = \(120 + 300 = 420\) kg, total sand = \(180 + 400 = 580\) kg, total gravel = \(300 + 500 = 800\) kg. The ratio of cement : sand : gravel is \(420 : 580 : 800\). Dividing by \(20\) gives \(21 : 29 : 40\). (c) Original cost of \(4.5\) tonnes = \(4.5 \times 120 = £540\). With a \(15\%\) increase, the new cost is \(540 \times 1.15 = £621\).

(a) M1 for \(750 \times \frac{2}{10}\). A1 for \(150\). (b) M1 for finding the masses of cement, sand, and gravel in either mix A or mix B. M1 for summing the respective masses to find the total mass of each ingredient in mix C. A1 for writing the ratio as \(420 : 580 : 800\). A1 for the simplest form \(21 : 29 : 40\). (c) M1 for finding the original cost of \(4.5\) tonnes (\(540\)) or the new cost per tonne (\(138\)). A1 for \(621\).

(a) \(f(4) = \frac{3(4) + 1}{4 - 2} = \frac{13}{2} = 6.5\). (b) Let \(y = \frac{3x + 1}{x - 2}\). Then \(y(x - 2) = 3x + 1 \implies yx - 2y = 3x + 1 \implies yx - 3x = 2y + 1 \implies x(y - 3) = 2y + 1 \implies x = \frac{2y + 1}{y - 3}\). Thus, \(f^{-1}(x) = \frac{2x + 1}{x - 3}\). (c) \(gf(x) = g\left(\frac{3x + 1}{x - 2}\right) = 2\left(\frac{3x + 1}{x - 2}\right) - 5 = \frac{6x + 2 - 5(x - 2)}{x - 2} = \frac{6x + 2 - 5x + 10}{x - 2} = \frac{x + 12}{x - 2}\). (d) \(gf(x) = 3 \implies \frac{x + 12}{x - 2} = 3 \implies x + 12 = 3x - 6 \implies 2x = 18 \implies x = 9\).

(a) B1 for \(6.5\) or \(\frac{13}{2}\). (b) M1 for setting \(y = \frac{3x+1}{x-2}\) and attempting to clear the fraction. M1 for grouping terms in \(x\) and factorising. A1 for \(f^{-1}(x) = \frac{2x+1}{x-3}\). (c) M1 for substituting \(f(x)\) into \(g(x)\) and writing with a common denominator. A1 for \(\frac{x+12}{x-2}\). (d) M1 for setting their expression for \(gf(x) = 3\) and attempting to solve for \(x\). A1 for \(x = 9\).

(a) Since \(AT\) is a tangent to the circle at \(A\), the angle between the tangent and the radius is \(90^\circ\), so angle \(OAT = 90^\circ\). In triangle \(OAT\), the angles sum to \(180^\circ\), so angle \(AOT = 180^\circ - 90^\circ - 36^\circ = 54^\circ\). (b) Since \(D, O, B, T\) is a straight line, angle \(AOB = \text{angle } AOT = 54^\circ\). In triangle \(OAB\), \(OA = OB\) because both are radii of the circle, meaning triangle \(OAB\) is an isosceles triangle. Thus, angle \(OAB = \text{angle } OBA = \frac{180^\circ - 54^\circ}{2} = 63^\circ\). (c) The angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at the circumference. Thus, angle \(ADB = \frac{1}{2} \times \text{angle } AOB = \frac{1}{2} \times 54^\circ = 27^\circ\). Alternatively, since \(BD\) is a diameter of the circle, the angle in a semicircle is a right angle, so angle \(DAB = 90^\circ\). Then, in triangle \(DAB\), the angles sum to \(180^\circ\), so angle \(ADB = 180^\circ - 90^\circ - 63^\circ = 27^\circ\).

(a) B1 for \(54^\circ\). B1 for stating that the angle between a tangent and a radius is \(90^\circ\). (b) M1 for recognizing that triangle \(OAB\) is isosceles because \(OA = OB\). M1 for calculating \(\frac{180^\circ - 54^\circ}{2}\). A1 for \(63^\circ\). (c) M1 for using the circle theorem that the angle at the centre is twice the angle at the circumference (or that the angle in a semicircle is \(90^\circ\) and using the angle sum of a triangle). A1 for \(27^\circ\). B1 for a correct geometric reason.