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Factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Factorise the denominator: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Divide both by the common factor \(2x + 1\) to obtain the simplified fraction: \(\frac{x - 3}{2x - 1}\).

M1 for factorising numerator: \((2x + 1)(x - 3)\); M1 for factorising denominator: \((2x - 1)(2x + 1)\); A1 for final answer \(\frac{x - 3}{2x - 1}\)

Let the original price be \(x\). A 15% reduction means the sale price is 85% of the original price. Therefore, \(0.85x = 459\), which gives \(x = \frac{459}{0.85} = 540\).

M1 for setting up \(0.85x = 459\) or \(\frac{459}{85} \times 100\); A1 for 540

The formula for the total surface area of a cone is \(A = \pi r^2 + \pi r l\), where \(r\) is the radius and \(l\) is the slant height. Substituting the given values: \(A = \pi(6)^2 + \pi(6)(10) = 36\pi + 60\pi = 96\pi\text{ cm}^2\).

M1 for correctly substituting into \(\pi r^2 + \pi r l\): \(\pi (6)^2 + \pi (6)(10)\); A1 for \(96\pi\)

Using the cosine rule: \(BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(BAC)\). Substituting the lengths: \(8^2 = 7^2 + 9^2 - 2(7)(9)\cos(A)\) which simplifies to \(64 = 49 + 81 - 126\cos(A)\). Then \(126\cos(A) = 130 - 64 = 66\), so \(\cos(A) = \frac{66}{126}\). Thus, \(A = \arccos\left(\frac{11}{21}\right) \approx 58.41^{\circ}\).

M1 for applying the cosine rule correctly: \(8^2 = 7^2 + 9^2 - 2(7)(9)\cos(A)\); M1 for rearranging to find \(\cos(A) = \frac{66}{126}\); A1 for \(58.4\) (or \(58.4^{\circ}\))

The probability of selecting two red marbles is \(P(\text{R, R}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}\). The probability of selecting two blue marbles is \(P(\text{B, B}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}\). The probability of selecting two marbles of the same color is \(P(\text{R, R}) + P(\text{B, B}) = \frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}\).

M1 for finding \(P(\text{R, R}) = \frac{20}{56}\) or \(P(\text{B, B}) = \frac{6}{56}\); M1 for adding the two probabilities: \(\frac{20}{56} + \frac{6}{56}\); A1 for \(\frac{13}{28}\) (or 3sf decimal equivalent \(0.464\))

Using Pythagoras' theorem, the initial height of the top of the ladder is \(h_1 = \sqrt{5^2 - 1.5^2} = \sqrt{22.75} \approx 4.7697\text{ m}\). After slipping down by \(0.8\text{ m}\), the new height is \(h_2 = 4.7697 - 0.8 = 3.9697\text{ m}\). The new distance of the foot of the ladder from the wall is \(d = \sqrt{5^2 - h_2^2} = \sqrt{25 - (3.9697)^2} = \sqrt{25 - 15.7585} = \sqrt{9.2415} \approx 3.04\text{ m}\).

M1 for finding the initial height: \(\sqrt{5^2 - 1.5^2} \approx 4.77\); M1 for finding the new distance using Pythagoras with the new height: \(\sqrt{5^2 - (4.77 - 0.8)^2}\); A1 for \(3.04\)

First, find the midpoints of each interval: \(5, 15, 25, 35\). Next, calculate the sum of the products of each midpoint and its corresponding frequency: \(8 \times 5 + 15 \times 15 + 11 \times 25 + 6 \times 35 = 40 + 225 + 275 + 210 = 750\). Divide by the total frequency (40): \(\text{Mean} = \frac{750}{40} = 18.75\text{ minutes}\).

M1 for calculating at least three correct midpoints \(5, 15, 25, 35\); M1 for calculating \(\sum f \cdot x = 750\); A1 for \(18.75\)

Comparing corresponding sides, the width of triangle \(A\) is \(3 - 1 = 2\) units, and the width of triangle \(B\) is \(-6 - (-2) = -4\) units. The ratio is \(\frac{-4}{2} = -2\), indicating a scale factor of \(-2\). The lines connecting corresponding vertices \((1,1)\) to \((-2,-2)\), \((3,1)\) to \((-6,-2)\), and \((1,4)\) to \((-2,-8)\) all intersect at the origin \((0,0)\). Therefore, the transformation is an enlargement with scale factor \(-2\) and center of enlargement \((0,0)\).

B1 for 'Enlargement'; B1 for scale factor \(-2\); B1 for center of enlargement \((0,0)\) or origin

First, factorise the numerator and the denominator:

Numerator: \(3x^2 - 14x - 5 = (3x + 1)(x - 5)\)
Denominator: \(x^2 - 25 = (x - 5)(x + 5)\)

Now rewrite the fraction:
\(\frac{(3x + 1)(x - 5)}{(x - 5)(x + 5)}\)

Cancel the common factor of \((x - 5)\):
\(\frac{3x + 1}{x + 5}\)

M1 for factorising the numerator to \((3x + 1)(x - 5)\)
M1 for factorising the denominator to \((x - 5)(x + 5)\)
A1 for final correct simplified fraction \(\frac{3x + 1}{x + 5}\)

The volume of a sphere is given by \(V_{\text{sphere}} = \frac{4}{3}\pi r^3\).
\(V_{\text{sphere}} = \frac{4}{3} \times \pi \times 4^3 = \frac{256}{3}\pi\text{ cm}^3\).

The volume of a cone is given by \(V_{\text{cone}} = \frac{1}{3}\pi R^2 h\).
\(V_{\text{cone}} = \frac{1}{3} \times \pi \times 3^2 \times h = 3\pi h\).

Since the volume remains the same:
\(3\pi h = \frac{256}{3}\pi\)
\(3h = \frac{256}{3}\)
\(h = \frac{256}{9} \approx 28.4\text{ cm}\).

M1 for correct volume of the sphere expression: \(\frac{256}{3}\pi\)
M1 for equating their sphere volume to the cone volume formula and attempting to solve for \(h\)
A1 for \(28.4\) (or \(256/9\))

Let \(\theta\) be the angle the ladder makes with the horizontal ground. The ladder represents the hypotenuse of a right-angled triangle, and the distance from the wall is the adjacent side.

Using the cosine ratio:
\(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{2.5}{6.5} = \frac{5}{13}\)

\(\theta = \cos^{-1}\left(\frac{5}{13}\right) \approx 67.38^\circ\)

To 1 decimal place, \(\theta = 67.4^\circ\).

M1 for setting up the correct trigonometric equation, e.g., \(\cos(\theta) = \frac{2.5}{6.5}\)
A1 for \(67.4\)

Find the midpoint of each interval:
- \(80 < m \le 100\): midpoint = \(90\)
- \(100 < m \le 120\): midpoint = \(110\)
- \(120 < m \le 140\): midpoint = \(130\)
- \(140 < m \le 160\): midpoint = \(150\)

Calculate the sum of \(\text{frequency} \times \text{midpoint}\):
\(\sum f x = (12 \times 90) + (18 \times 110) + (15 \times 130) + (5 \times 150)\)
\(\sum f x = 1080 + 1980 + 1950 + 750 = 5760\)

Divide the sum by the total frequency:
\(\text{Estimate of mean} = \frac{5760}{50} = 115.2\text{ g}\).

M1 for finding the correct midpoints (90, 110, 130, 150)
M1 for finding \(\sum f x = 5760\)
A1 for \(115.2\)

The discounted price represents \(100\% - 15\% = 85\%\) of the original price.

Let \(P\) be the original price:
\(0.85 \times P = 357\)

\(P = \frac{357}{0.85} = 420\)

Therefore, the original price was \(\$420\).

M1 for setting up the equation \(0.85 \times P = 357\) or equivalent division \(\frac{357}{0.85}\)
A1 for \(420\)

Using the cosine rule:
\(BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(BAC)\)
\(BC^2 = 8.4^2 + 6.5^2 - 2 \times 8.4 \times 6.5 \times \cos(42^\circ)\)
\(BC^2 = 70.56 + 42.25 - 109.2 \times \cos(42^\circ)\)
\(BC^2 \approx 112.81 - 109.2 \times 0.74314...\)
\(BC^2 \approx 112.81 - 81.1514...\)
\(BC^2 \approx 31.658...\)
\(BC = \sqrt{31.658...} \approx 5.6265...\text{ cm}\)

To 3 significant figures, \(BC = 5.63\text{ cm}\).

M1 for substituting correctly into the cosine rule: \(8.4^2 + 6.5^2 - 2 \times 8.4 \times 6.5 \times \cos(42^\circ)\)
A1 for intermediate value \(BC^2 \approx 31.66\) or \(BC = \sqrt{31.66}\)
A1 for \(5.63\) (allow answers in the range \(5.62\) to \(5.63\))

Let us check the coordinates of mapped vertices:
- \((1, 1) \to (-1, -1)\)
- \((3, 1) \to (-1, -3)\)
- \((3, 4) \to (-4, -3)\)

We see that for each point \((x, y)\), the image point is \((-y, -x)\).
This is the coordinate mapping rule for a reflection in the line \(y = -x\).

M1 for identifying the transformation as a 'Reflection'
A1 for identifying the line of reflection as \(y = -x\)

The total number of beads is \(5 + 3 = 8\).

Probability of selecting two red beads:
\(P(\text{Red, Red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}\)

Probability of selecting two blue beads:
\(P(\text{Blue, Blue}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}\)

Probability that both beads are the same colour:
\(P(\text{Same Colour}) = P(\text{Red, Red}) + P(\text{Blue, Blue}) = \frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}\).

M1 for \(\frac{5}{8} \times \frac{4}{7}\) or \(\frac{3}{8} \times \frac{2}{7}\)
M1 for addition of their two calculated probabilities
A1 for \(13/28\)

First, factorise the numerator by splitting the middle term or by inspection: \(2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\). Next, factorise the denominator as a difference of two squares: \(4x^2 - 1 = (2x)^2 - 1^2 = (2x + 1)(2x - 1)\). Now, divide the numerator and the denominator by their common factor, \(2x + 1\): \(\frac{(2x+1)(x-3)}{(2x+1)(2x-1)} = \frac{x-3}{2x-1}\).

M1 for factorising the numerator correctly: \((2x + 1)(x - 3)\)
M1 for factorising the denominator correctly: \((2x + 1)(2x - 1)\)
A0.7 for the final simplified fraction \(\frac{x-3}{2x-1}\)

The volume of a sphere of radius \(r\) is given by the formula \(V_{\text{sphere}} = \frac{4}{3}\pi r^3\). The volume of a cylinder of radius \(r\) and height \(h = 2r\) is given by \(V_{\text{cylinder}} = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3\). To find the ratio of the volume of the sphere to the volume of the cylinder, we write: \(\frac{4}{3}\pi r^3 : 2\pi r^3\). Dividing both parts of the ratio by \(\pi r^3\) gives: \(\frac{4}{3} : 2\). Multiplying both parts by 3 to clear the fraction gives: \(4 : 6\). Simplifying this ratio to its lowest terms gives: \(2 : 3\).

M1 for expressing cylinder volume as \(2\pi r^3\) or sphere volume as \(\frac{4}{3}\pi r^3\)
M1 for setting up ratio \(\frac{4}{3}\pi r^3 : 2\pi r^3\) and attempting to simplify
A0.7 for \(2:3\)

Using the Cosine Rule to find the angle \(ABC\) (which we can call angle \(B\)), we have: \(b^2 = a^2 + c^2 - 2ac \cos(B)\). Here, \(b = AC = 13\), \(a = BC = 8\), and \(c = AB = 7\). Substituting these values into the formula: \(13^2 = 8^2 + 7^2 - 2(8)(7) \cos(B)\) which simplifies to \(169 = 64 + 49 - 112 \cos(B)\). Simplify further: \(169 = 113 - 112 \cos(B)\). Rearranging to solve for \(\cos(B)\): \(112 \cos(B) = 113 - 169\), which gives \(112 \cos(B) = -56\). Thus, \(\cos(B) = -\frac{56}{112} = -0.5\). Calculating the inverse cosine: \(B = \arccos(-0.5) = 120^\circ\).

M1 for substituting correctly into the Cosine Rule: \(13^2 = 8^2 + 7^2 - 2(8)(7)\cos(B)\)
M1 for reaching \(\cos(B) = -0.5\)
A0.7 for \(120\)

The total number of beads initially is \(5 + 3 = 8\). We want to find the probability of selecting beads of different colours, which can happen in two ways: (Red, Blue) or (Blue, Red). First, the probability of selecting a Red bead then a Blue bead: \(P(\text{Red, Blue}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\). Second, the probability of selecting a Blue bead then a Red bead: \(P(\text{Blue, Red}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\). Since these two outcomes are mutually exclusive, we add their probabilities to find the total probability: \(P(\text{different colours}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\) (or approximately \(0.536\)).

M1 for calculating the probability of one combination, e.g., \(\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)
M1 for adding the two correct combinations: \(\frac{15}{56} + \frac{15}{56}\)
A0.7 for \(\frac{15}{28}\) (or equivalent fraction/decimal)

The reduced price of \(\$578\) represents \(100\% - 15\% = 85\%\) of the original price. Let \(x\) represent the original price. We can write the equation: \(0.85x = 578\). To find \(x\), divide both sides by \(0.85\): \(x = \frac{578}{0.85} = 680\). Thus, the original price of the laptop was \(\$680\).

M1 for equating \(85\%\) to \(\$578\) or writing \(578 \div 0.85\)
A1.7 for \(680\)

When a point \((x, y)\) is reflected in the line \(y = x\), its coordinates swap places to become \((y, x)\). Reflecting \(P(3, -2)\) gives \(Q(-2, 3)\). Next, we translate \(Q(-2, 3)\) by the vector \(\begin{pmatrix} -4 \\ 5 \end{pmatrix}\). The new coordinates are: \(x = -2 + (-4) = -6\) and \(y = 3 + 5 = 8\). Therefore, the coordinates of \(R\) are \((-6, 8)\).

M1 for finding the correct coordinates of \(Q = (-2, 3)\)
M1 for adding the translation vector components to their \(Q\) coordinates
A0.7 for \((-6, 8)\)

The sum of the initial set of 5 numbers is \(5 \times 12 = 60\). The sum of the new set of 6 numbers is \(6 \times 13.5 = 81\). The value of the 6th number is the difference between these two sums: \(81 - 60 = 21\).

M1 for calculating the sum of either the first 5 numbers (\(60\)) or all 6 numbers (\(81\))
M1 for subtracting the sum of 5 numbers from the sum of 6 numbers: \(81 - 60\)
A0.7 for \(21\)

The ladder, the ground, and the wall form a right-angled triangle. Let \(h\) be the height up the wall reached by the ladder. By Pythagoras' theorem, we have: \(h^2 + 2.5^2 = 6.5^2\). Substituting the squared terms: \(h^2 + 6.25 = 42.25\). Rearranging gives \(h^2 = 42.25 - 6.25 = 36\). Taking the square root: \(h = \sqrt{36} = 6\) metres.

M1 for setting up Pythagoras' theorem: \(h^2 + 2.5^2 = 6.5^2\) or \(h = \sqrt{6.5^2 - 2.5^2}\)
M1 for evaluating \(h^2 = 36\)
A0.7 for \(6\)

First, factorise the numerator by factoring out the common term and then using the difference of two squares:
\(2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2)\)

Next, factorise the quadratic expression in the denominator:
\(x^2 - x - 6 = (x - 3)(x + 2)\)

Substitute the factorised terms back into the fraction:
\(\frac{2(x - 2)(x + 2)}{(x - 3)(x + 2)}\)

Cancel the common factor of \(x + 2\) from the numerator and the denominator:
\(\frac{2(x - 2)}{x - 3}\) or \(\frac{2x - 4}{x - 3}\)

M1 for factorising the numerator to \(2(x-2)(x+2)\)
M1 for factorising the denominator to \((x-3)(x+2)\)
A0.7 for the final answer \(\frac{2(x-2)}{x-3}\) or \(\frac{2x-4}{x-3}\)

Let \(V\) be the initial value of the clock.

After a 15% increase in the first year, the value becomes:
\(V \times 1.15\)

After an 8% increase in the second year, the value becomes:
\(V \times 1.15 \times 1.08 = V \times 1.242\)

We are given that the final value is $1863:
\(1.242V = 1863\)

Solve for \(V\):
\(V = \frac{1863}{1.242} = 1500\)

The initial value of the clock was $1500.

M1 for a correct product of multipliers, e.g., \(1.15 \times 1.08\) or \(1.242\)
M1 for setting up the equation \(V \times 1.242 = 1863\) or for finding the intermediate value \(\frac{1863}{1.08} = 1725\)
A0.7 for the final answer of 1500

**(a)**
Factorise the numerator:
$$2x^2 - 5x - 3 = (2x + 1)(x - 3)$$
Factorise the denominator as a difference of two squares:
$$4x^2 - 1 = (2x + 1)(2x - 1)$$
Simplify by cancelling the common factor $(2x + 1)$:
$$\frac{(2x + 1)(x - 3)}{(2x + 1)(2x - 1)} = \frac{x - 3}{2x - 1}$$

**(b)**
Multiply each term by the common denominator $(y+2)(2y-1)$:
$$3(2y - 1) + 5(y + 2) = 2(y + 2)(2y - 1)$$
$$6y - 3 + 5y + 10 = 2(2y^2 + 3y - 2)$$
$$11y + 7 = 4y^2 + 6y - 4$$
Rearrange to form a quadratic equation equal to zero:
$$4y^2 - 5y - 11 = 0$$
Apply the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(4)(-11)}}{2(4)}$$
$$y = \frac{5 \pm \sqrt{25 + 176}}{8} = \frac{5 \pm \sqrt{201}}{8}$$
$$y_1 = \frac{5 + 14.177}{8} \approx 2.40$$
$$y_2 = \frac{5 - 14.177}{8} \approx -1.15$$

**(c)**
Find a common denominator:
$$\frac{4(x+1) - 3(x-3)}{(x-3)(x+1)}$$
$$= \frac{4x + 4 - 3x + 9}{(x-3)(x+1)}$$
$$= \frac{x + 13}{(x-3)(x+1)}$$

**(a)** [4 marks]
- M1 for factorising numerator: $(2x + 1)(x - 3)$
- M1 for factorising denominator: $(2x + 1)(2x - 1)$
- A1 for cancelling common factor $(2x+1)$
- A1 for final answer $\frac{x - 3}{2x - 1}$

**(b)** [5 marks]
- M1 for clearing fractions: $3(2y - 1) + 5(y + 2) = 2(y + 2)(2y - 1)$
- M1 for expanding correctly to form $4y^2 - 5y - 11 = 0$
- M1 for applying the quadratic formula with their $a, b, c$ values
- A1 for $y = 2.40$ (or $2.4$)
- A1 for $y = -1.15$

**(c)** [2.8 marks]
- M1 for common denominator $(x-3)(x+1)$
- M1 for expanding numerator $4(x+1) - 3(x-3)$
- A0.8 for final answer $\frac{x + 13}{(x-3)(x+1)}$ or $\frac{x+13}{x^2-2x-3}$

**(a)**
First, find the slant height, $l$, of the cone using Pythagoras' theorem:
$$l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\text{ cm}$$
The curved surface area of the cone is:
$$\text{CSA} = \pi r l = \pi \times 6 \times 10 = 60\pi\text{ cm}^2$$
The area of the circular base is:
$$\text{Base Area} = \pi r^2 = \pi \times 6^2 = 36\pi\text{ cm}^2$$
Total surface area of the cone is:
$$\text{Total Area} = 60\pi + 36\pi = 96\pi\text{ cm}^2$$

**(b)**
First, calculate the volume of the cone:
$$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 6^2 \times 8 = 96\pi\text{ cm}^3$$
Since the cone is recast into 5 identical spheres, the volume of one sphere is:
$$V_{\text{sphere}} = \frac{96\pi}{5} = 19.2\pi\text{ cm}^3$$
The formula for the volume of a sphere is $V = \frac{4}{3}\pi R^3$. Equating these:
$$\frac{4}{3}\pi R^3 = 19.2\pi$$
$$\frac{4}{3} R^3 = 19.2$$
$$R^3 = 19.2 \times \frac{3}{4} = 14.4$$
$$R = \sqrt[3]{14.4} \approx 2.43288...\text{ cm}$$
To 3 significant figures, $R = 2.43\text{ cm}$.

**(c)**
The surface area of a sphere is given by the formula $A = 4\pi R^2$:
$$A = 4\pi \times (2.43288)^2$$
$$A = 4\pi \times 5.9189$$
$$A \approx 74.379\text{ cm}^2$$
To 3 significant figures, $A = 74.4\text{ cm}^2$.

**(a)** [3.8 marks]
- M1 for finding slant height $l = \sqrt{6^2 + 8^2} = 10$
- M1 for curved surface area formula $\pi \times 6 \times 10 = 60\pi$
- M1 for base area formula $\pi \times 6^2 = 36\pi$
- A0.8 for summing to $96\pi$ with no errors shown

**(b)** [4 marks]
- M1 for finding volume of the cone $V = 96\pi$ (or approx 301.6)
- M1 for setting $\frac{4}{3}\pi R^3 = \frac{\text{their } V}{5}$
- M1 for isolating $R^3 = 14.4$
- A1 for $2.43$ (accept 2.43 to 2.44)

**(c)** [4 marks]
- M1 for recalling formula $A = 4\pi R^2$
- M1 for substituting their radius $R$ into the formula
- A1 for $74.4$ (accept 74.3 to 74.5)

**(a)**
Using the Cosine Rule to find $AC$:
$$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$$
$$AC^2 = 75^2 + 110^2 - 2(75)(110)\cos(54^\circ)$$
$$AC^2 = 5625 + 12100 - 16500\cos(54^\circ)$$
$$AC^2 = 17725 - 16500(0.587785)$$
$$AC^2 = 17725 - 9698.46 = 8026.54$$
$$AC = \sqrt{8026.54} \approx 89.591\text{ m}$$
To 3 significant figures, $AC = 89.6\text{ m}$.

**(b)**
Using the Sine Rule to find angle $ACB$:
$$\frac{\sin(\angle ACB)}{AB} = \frac{\sin(\angle ABC)}{AC}$$
$$\frac{\sin(\angle ACB)}{75} = \frac{\sin(54^\circ)}{89.591}$$
$$\sin(\angle ACB) = \frac{75 \times \sin(54^\circ)}{89.591}$$
$$\sin(\angle ACB) = \frac{75 \times 0.809017}{89.591} \approx 0.67726$$
$$\angle ACB = \sin^{-1}(0.67726) \approx 42.63^\circ$$
To 1 decimal place, $\angle ACB = 42.6^\circ$.

**(c)**
Using the area of a triangle formula:
$$\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\angle ABC)$$
$$\text{Area} = \frac{1}{2} \times 75 \times 110 \times \sin(54^\circ)$$
$$\text{Area} = 4125 \times 0.809017 \approx 3337.19\text{ m}^2$$
To 3 significant figures, $\text{Area} = 3340\text{ m}^2$.

**(a)** [4.8 marks]
- M1 for writing down correct cosine rule formula
- M1 for correct substitution: $75^2 + 110^2 - 2(75)(110)\cos(54^\circ)$
- A1 for $8026.5$ or better
- A1.8 for $89.6$ (accept 89.5 to 89.7)

**(b)** [4 marks]
- M1 for correct sine rule formula
- M1 for correct substitution using their $AC$ value: $\frac{\sin(\angle ACB)}{75} = \frac{\sin(54^\circ)}{89.6}$
- A1 for $\sin(\angle ACB) \approx 0.677$
- A1 for $42.6^\circ$ (accept 42.5 to 42.7)

**(c)** [3 marks]
- M1 for correct area formula: $\frac{1}{2} a c \sin B$
- M1 for correct substitution: $\frac{1}{2} \times 75 \times 110 \times \sin(54^\circ)$
- A1 for 3340 (accept 3330 to 3340)

**(a)(i)**
Using the compound interest formula $A = P\left(1 + \frac{r}{100}\right)^t$:
$$A = 6500 \times (1.024)^5$$
$$A \approx 6500 \times 1.1258999$$
$$A \approx 7318.35$$
To the nearest dollar, the value is $7318.

**(a)(ii)**
We want to find the smallest integer $n$ such that:
$$6500 \times (1.024)^n > 9000$$
$$(1.024)^n > \frac{9000}{6500}$$
$$(1.024)^n > 1.3846$$
Taking logarithms on both sides:
$$n \log(1.024) > \log(1.3846)$$
$$n > \frac{\log(1.3846)}{\log(1.024)}$$
$$n > \frac{0.14133}{0.01030} \approx 13.72\text{ years}$$
Since $n$ must be a complete year, it takes 14 complete years.

**(b)**
Let $P$ be the original price of the computer.
$$P \times (1 + 0.15) = 828$$
$$1.15 P = 828$$
$$P = \frac{828}{1.15} = 720$$
The price before sales tax was $720.

**(c)**
Let the original value of the car be $V_0$.
Each year, the value is multiplied by $(1 - 0.12) = 0.88$.
After 4 years, the remaining value is:
$$V = V_0 \times (0.88)^4$$
$$(0.88)^4 = 0.599695$$
Expressing this as a percentage: $0.599695 \times 100\% \approx 59.97\%$.
To 3 significant figures, the remaining value is $60.0\%$.

**(a)(i)** [3 marks]
- M1 for $6500 \times 1.024^5$
- A1 for 7318.35
- A1 for 7318 (nearest dollar)

**(a)(ii)** [3.8 marks]
- M1 for writing the inequality or equation $6500 \times 1.024^n > 9000$
- M1 for trial and error showing values for $n=13$ ($8849.63$) and $n=14$ ($9062.02$), OR for using logarithms correctly
- A1.8 for 14

**(b)** [3 marks]
- M1 for $P \times 1.15 = 828$
- M1 for division $\frac{828}{1.15}$
- A1 for 720

**(c)** [2 marks]
- M1 for calculation $0.88^4$
- A1 for $60.0$ or $59.97$

**(a)**
Reflection in the line $y = -x$ maps any point $(x, y)$ to $(-y, -x)$.
Applying this rule to each vertex:
- $A(1, 2) \to A'(-2, -1)$
- $B(4, 2) \to B'(-2, -4)$
- $C(3, 4) \to C'(-4, -3)$

**(b)**
An enlargement with scale factor $2.5$ and center $(0,0)$ multiplies each coordinate by $2.5$:
- $A(1, 2) \to A''(2.5, 5)$
- $B(4, 2) \to B''(10, 5)$
- $C(3, 4) \to C''(7.5, 10)$

**(c)**
A shear with the $x$-axis invariant and shear factor $k=2$ maps $(x, y)$ to $(x + ky, y)$, which is $(x + 2y, y)$.
- For $A(1, 2)$: $(1 + 2(2), 2) = (5, 2)$
- For $B(4, 2)$: $(4 + 2(2), 2) = (8, 2)$
- For $C(3, 4)$: $(3 + 2(4), 4) = (11, 4)$
So the vertices of $W$ are $(5, 2)$, $(8, 2)$, and $(11, 4)$.

**(d)**
The standard matrix for a shear with the $x$-axis invariant and shear factor $k$ is:
$$\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$$
Since the shear factor is $2$, the matrix is:
$$\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$$
We can verify this with $A(1, 2)$:
$$\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1(1) + 2(2) \\ 0(1) + 1(2) \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}$$

**(a)** [3 marks]
- B1 for each correct point: $A'(-2, -1)$, $B'(-2, -4)$, $C'(-4, -3)$

**(b)** [3 marks]
- B1 for each correct point: $A''(2.5, 5)$, $B''(10, 5)$, $C''(7.5, 10)$

**(c)** [3 marks]
- M1 for using transformation rule $(x+2y, y)$
- A2 for all three points correct: $(5, 2)$, $(8, 2)$, and $(11, 4)$ (A1 if one error)

**(d)** [2.8 marks]
- M1 for writing a matrix of the form $\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$
- A1.8 for the correct matrix $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$

**(a)**
The base $ABCD$ is a rectangle. Use Pythagoras' theorem in the right-angled triangle $ABC$:
$$AC^2 = AB^2 + BC^2$$
$$AC^2 = 12^2 + 5^2 = 144 + 25 = 169$$
$$AC = \sqrt{169} = 13\text{ cm}$$

**(b)**
The vertical edge $CG$ is perpendicular to the base $ABCD$, and therefore perpendicular to the line $AC$.
In the right-angled triangle $ACG$, use Pythagoras' theorem:
$$AG^2 = AC^2 + CG^2$$
$$AG^2 = 13^2 + 8^2 = 169 + 64 = 233$$
$$AG = \sqrt{233} \approx 15.264\text{ cm}$$
To 3 significant figures, $AG = 15.3\text{ cm}$.

**(c)**
The angle that $AG$ makes with the horizontal base $ABCD$ is the angle $\angle CAG$.
In the right-angled triangle $ACG$:
$$\tan(\angle CAG) = \frac{CG}{AC} = \frac{8}{13}$$
$$\angle CAG = \tan^{-1}\left(\frac{8}{13}\right) \approx \tan^{-1}(0.61538) \approx 31.607^\circ$$
To 1 decimal place, the angle is $31.6^\circ$.

**(d)**
The angle that $AG$ makes with the vertical edge $AE$ is the angle $\angle EAG$.
In the right-angled triangle $AEG$ (where $EG$ is the diagonal of the top face, which is equal to $AC = 13\text{ cm}$):
$$\tan(\angle EAG) = \frac{EG}{AE} = \frac{13}{8}$$
$$\angle EAG = \tan^{-1}\left(\frac{13}{8}\right) = \tan^{-1}(1.625) \approx 58.39^\circ$$
To 1 decimal place, the angle is $58.4^\circ$.

**(a)** [3 marks]
- M1 for $12^2 + 5^2$
- A1 for $\sqrt{169}$
- A1 for 13

**(b)** [3 marks]
- M1 for $AC^2 + 8^2$ (or $12^2 + 5^2 + 8^2$)
- A1 for $\sqrt{233}$
- A1 for 15.3

**(c)** [3 marks]
- M1 for identifying angle $\angle CAG$ and using tangent (or sine/cosine with their $AG$)
- M1 for $\tan(\angle CAG) = \frac{8}{13}$
- A1 for $31.6^\circ$ (accept 31.5 to 31.7)

**(d)** [2.8 marks]
- M1 for identifying angle $\angle EAG$ and using $\tan(\angle EAG) = \frac{13}{8}$ or alternate valid trig ratio
- A1.8 for $58.4^\circ$ (accept 58.3 to 58.5)

**(a)(i)**
Total number of pens = $6 + 4 + 2 = 12$.
Probability of selecting a red pen first is $\frac{6}{12} = \frac{1}{2}$.
Since the selection is without replacement, there are now 5 red pens and 11 total pens.
Probability of selecting a second red pen is $\frac{5}{11}$.
$$P(\text{both red}) = \frac{6}{12} \times \frac{5}{11} = \frac{30}{132} = \frac{5}{22} \approx 0.227$$

**(a)(ii)**
It is easier to first find the probability that both pens are of the same color.
$$P(\text{same color}) = P(\text{both red}) + P(\text{both blue}) + P(\text{both green})$$
- $P(\text{both red}) = \frac{6}{12} \times \frac{5}{11} = \frac{30}{132}$
- $P(\text{both blue}) = \frac{4}{12} \times \frac{3}{11} = \frac{12}{132}$
- $P(\text{both green}) = \frac{2}{12} \times \frac{1}{11} = \frac{2}{132}$
$$P(\text{same color}) = \frac{30}{132} + \frac{12}{132} + \frac{2}{132} = \frac{44}{132} = \frac{1}{3}$$
Since the two pens are either the same color or different colors:
$$P(\text{different colors}) = 1 - P(\text{same color}) = 1 - \frac{1}{3} = \frac{2}{3} \approx 0.667$$

**(b)**
Since the pens are selected with replacement, the probability of selecting a green pen in any draw is constant:
$$P(\text{Green}) = \frac{2}{12} = \frac{1}{6}$$
Thus, the probability of NOT selecting a green pen in any draw is:
$$P(\text{Not Green}) = 1 - \frac{1}{6} = \frac{5}{6}$$
The probability of getting NO green pens in three selections is:
$$P(\text{no green}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$$
The probability of getting at least one green pen is:
$$P(\text{at least one green}) = 1 - P(\text{no green}) = 1 - \frac{125}{216} = \frac{91}{216} \approx 0.421$$

**(a)(i)** [3 marks]
- M1 for product of two fractions, second denominator being 11
- M1 for $\frac{6}{12} \times \frac{5}{11}$ or equivalent
- A1 for $\frac{5}{22}$ or $0.227$ (accept $0.227$ to $0.23$)

**(a)(ii)** [4.8 marks]
- M1 for finding at least one other same-color pair: e.g., $P(\text{blue, blue}) = \frac{4}{12} \times \frac{3}{11}$
- M1 for summing same-color probabilities: $P(R,R) + P(B,B) + P(G,G)$
- M1 for subtracting their sum from 1
- A1.8 for $\frac{2}{3}$ or $0.667$ (accept $0.667$ to $0.67$)

**(b)** [4 marks]
- M1 for identifying the probability of not green on one draw is $\frac{10}{12}$ or $\frac{5}{6}$
- M1 for calculating $(\text{not green})^3 = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$
- M1 for $1 - \left(\frac{5}{6}\right)^3$
- A1 for $\frac{91}{216}$ or $0.421$ (accept $0.421$ to $0.422$)

**(a)**
Differentiate the curve $y = 2x^3 - 3x^2 - 12x + 5$ term-by-term with respect to $x$:
$$\frac{dy}{dx} = 3(2x^2) - 2(3x^1) - 12 + 0$$
$$\frac{dy}{dx} = 6x^2 - 6x - 12$$

**(b)**
At a stationary point, the gradient $\frac{dy}{dx} = 0$:
$$6x^2 - 6x - 12 = 0$$
Divide the entire equation by 6:
$$x^2 - x - 2 = 0$$
Factorise the quadratic:
$$(x - 2)(x + 1) = 0$$
So, the $x$-coordinates of the stationary points are $x = 2$ and $x = -1$.

Find the corresponding $y$-coordinates by substituting the $x$-values back into the original curve equation:
- For $x = 2$:
$$y = 2(2)^3 - 3(2)^2 - 12(2) + 5$$
$$y = 2(8) - 3(4) - 24 + 5 = 16 - 12 - 24 + 5 = -15$$
So, one stationary point is $(2, -15)$.

- For $x = -1$:
$$y = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5$$
$$y = 2(-1) - 3(1) + 12 + 5 = -2 - 3 + 12 + 5 = 12$$
So, the other stationary point is $(-1, 12)$.

**(c)**
To find the nature of each stationary point, we can use the second derivative test.
Differentiate $\frac{dy}{dx} = 6x^2 - 6x - 12$ with respect to $x$:
$$\frac{d^2y}{dx^2} = 12x - 6$$

Now test each stationary point:
- For $x = 2$:
$$\frac{d^2y}{dx^2} = 12(2) - 6 = 24 - 6 = 18$$
Since $\frac{d^2y}{dx^2} > 0$, the point $(2, -15)$ is a **local minimum**.

- For $x = -1$:
$$\frac{d^2y}{dx^2} = 12(-1) - 6 = -12 - 6 = -18$$
Since $\frac{d^2y}{dx^2} < 0$, the point $(-1, 12)$ is a **local maximum**.

**(a)** [2 marks]
- M1 for attempting to differentiate at least one term (e.g., $6x^2$ or $-6x$)
- A1 for $6x^2 - 6x - 12$

**(b)** [5 marks]
- M1 for setting their $\frac{dy}{dx} = 0$
- M1 for factorising or solving $x^2 - x - 2 = 0$ to get $x = 2, -1$
- M1 for substituting $x = 2$ or $x = -1$ into original equation to find $y$
- A1 for $(2, -15)$
- A1 for $(-1, 12)$

**(c)** [4.8 marks]
- M1 for finding the second derivative: $12x - 6$
- M1 for substituting $x = 2$ into their $\frac{d^2y}{dx^2}$ and concluding local minimum because it is positive
- M1 for substituting $x = -1$ into their $\frac{d^2y}{dx^2}$ and concluding local maximum because it is negative
- A1.8 for fully correct working and conclusions

**(a) (i)**
Using Pythagoras' theorem in the right-angled triangle formed by the radius, height, and slant height of the cone:
\(h_{\text{cone}} = \sqrt{l^2 - r^2}\)
\(h_{\text{cone}} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}\).

**(a) (ii)**
Total volume, \(V = V_{\text{cylinder}} + V_{\text{cone}}\)
\(V = \pi r^2 h_{\text{cylinder}} + \frac{1}{3}\pi r^2 h_{\text{cone}}\)
\(V = \pi \times 5^2 \times 12 + \frac{1}{3}\pi \times 5^2 \times 12\)
\(V = 300\pi + 100\pi = 400\pi \approx 1256.6\text{ cm}^3 \approx 1260\text{ cm}^3\) (to 3 s.f.).

**(a) (iii)**
Total surface area, \(A = A_{\text{base}} + A_{\text{curved, cylinder}} + A_{\text{curved, cone}}\)
\(A = \pi r^2 + 2\pi r h_{\text{cylinder}} + \pi r l\)
\(A = \pi \times 5^2 + 2\pi \times 5 \times 12 + \pi \times 5 \times 13\)
\(A = 25\pi + 120\pi + 65\pi = 210\pi \approx 659.7\text{ cm}^2 \approx 660\text{ cm}^2\) (to 3 s.f.).

**(b)**
Volume of one sphere:
\(V_{\text{sphere}} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \times 1.5^3 = 4.5\pi \approx 14.137\text{ cm}^3\).
Number of spheres \(N = \frac{400\pi}{4.5\pi} = \frac{400}{4.5} \approx 88.89\).
Maximum number of complete spheres is \(88\).

**(a)(i)**
M1: For using Pythagoras' theorem: \(\sqrt{13^2 - 5^2}\) or \(h^2 + 5^2 = 13^2\)
A1: Correctly showing 12

**(a)(ii)**
M1: For volume of cylinder: \(\pi \times 5^2 \times 12\) [\(300\pi\)]
M1: For volume of cone: \(\frac{1}{3} \pi \times 5^2 \times 12\) [\(100\pi\)]
A1: For 1260 or 1256.6 to 1257

**(a)(iii)**
M1: For area of base: \(\pi \times 5^2\) [\(25\pi\)]
M1: For curved area of cylinder: \(2 \pi \times 5 \times 12\) [\(120\pi\)]
M1: For curved area of cone: \(\pi \times 5 \times 13\) [\(65\pi\)]
A1: For 660 or 659.7 to 660

**(b)**
M1: For volume of one sphere: \(\frac{4}{3} \pi \times 1.5^3\) [\(4.5\pi\) or 14.1 to 14.14]
M1: For dividing total volume by volume of one sphere
A1: For 88 (must be integer)

**(a)**
Time \(= \frac{\text{Distance}}{\text{Speed}} = \frac{36}{x}\) hours.

**(b)**
New speed \(= x + 3\text{ km/h}\).
Time taken \(= \frac{36}{x+3}\) hours.

**(c) (i)**
\(24\text{ minutes} = \frac{24}{60}\text{ hours} = \frac{2}{5}\text{ hours} = 0.4\text{ hours}\).
Therefore, the equation is:
\[\frac{36}{x} - \frac{36}{x+3} = \frac{2}{5}\]
Multiply the entire equation by \(5x(x+3)\) to clear the denominators:
\[5 \times 36(x+3) - 5 \times 36x = 2x(x+3)\]
\[180(x+3) - 180x = 2x^2 + 6x\]
\[180x + 540 - 180x = 2x^2 + 6x\]
\[540 = 2x^2 + 6x\]
Divide the entire equation by 2:
\[270 = x^2 + 3x\]
Rearranging gives:
\[x^2 + 3x - 270 = 0\]

**(c) (ii)**
Factoring \(x^2 + 3x - 270 = 0\):
We look for two numbers that multiply to \(-270\) and add to \(3\). These are \(18\) and \(-15\).
\[(x + 18)(x - 15) = 0\]
Thus, \(x = 15\) or \(x = -18\).

**(c) (iii)**
Since speed cannot be negative, \(x = 15\).
Outward time \(= \frac{36}{15} = 2.4\text{ hours} = 2\text{ hours and } 24\text{ minutes}\).
Return time \(= \frac{36}{15+3} = \frac{36}{18} = 2\text{ hours}\).
Total time taken \(= 2.4 + 2 = 4.4\text{ hours} = 4\text{ hours and } 24\text{ minutes}\).

**(a)**
B1: For \(\frac{36}{x}\)

**(b)**
B1: For \(\frac{36}{x+3}\)

**(c)(i)**
M1: For converting 24 minutes to hours: \(\frac{24}{60}\) or \(0.4\) or \(\frac{2}{5}\)
M1: For setting up equation: \(\frac{36}{x} - \frac{36}{x+3} = \frac{24}{60}\) (or equivalent)
M1: For multiplying by common denominator to clear fractions: \(5 \times 36(x+3) - 180x = 2x(x+3)\) (or equivalent)
A1: For completing the simplification to show \(x^2 + 3x - 270 = 0\) with no errors

**(c)(ii)**
M1: For factoring \((x-15)(x+18)\) or correct use of quadratic formula with at most one sign error
A1: For \(x = 15\)
A1: For \(x = -18\)

**(c)(iii)**
M1: For substituting their positive \(x\) value into their time expressions
M1: For adding both times: \(2.4 + 2\)
A1: For \(4\text{ hours } 24\text{ minutes}\)

**(a)**
First, find the angle \(\angle BAC\):
\(\angle BAC = 140^\circ - 75^\circ = 65^\circ\).
Using the Cosine Rule to find \(BC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC)\)
\(BC^2 = 18^2 + 25^2 - 2(18)(25)\cos(65^\circ)\)
\(BC^2 = 324 + 625 - 900\cos(65^\circ)\)
\(BC^2 = 949 - 900 \times 0.422618 = 949 - 380.356 = 568.644\)
\(BC = \sqrt{568.644} \approx 23.846 \approx 23.8\text{ km}\) (to 3 s.f.).

**(b)**
Using the Sine Rule to find \(\angle ABC\):
\(\frac{\sin(\angle ABC)}{AC} = \frac{\sin(\angle BAC)}{BC}\)
\(\sin(\angle ABC) = \frac{25 \sin(65^\circ)}{23.846} \approx \frac{22.6577}{23.846} \approx 0.95017\)
\(\angle ABC = \sin^{-1}(0.95017) \approx 71.84^\circ\).
Let's determine the bearing of \(C\) from \(B\).
Draw a North-South line at both \(A\) and \(B\).
The angle between the South line at \(B\) and the line \(BA\) is \(75^\circ\) (alternate angles to the North-bearing line at \(A\)).
Since the angle \(\angle ABC = 71.84^\circ\) is less than \(75^\circ\), the line \(BC\) is to the right of the South line by \(75^\circ - 71.84^\circ = 3.16^\circ\) (going west of south).
More precisely, the South direction is at \(180^\circ\). The direction of \(BA\) from \(B\) is \(75^\circ + 180^\circ = 255^\circ\).
Subtracting \(\angle ABC = 71.84^\circ\) from \(255^\circ\) gives:
Bearing of \(C\) from \(B = 255^\circ - 71.84^\circ = 183.16^\circ \approx 183.2^\circ\) (or \(183^\circ\) to 3 s.f.).

**(c)**
The shortest distance \(d\) from \(A\) to the line \(BC\) is the perpendicular distance from \(A\) to \(BC\).
We can find this using the area of triangle \(ABC\):
\(\text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\angle BAC) = \frac{1}{2} \times 18 \times 25 \times \sin(65^\circ) \approx 203.92\text{ km}^2\).
Also,
\(\text{Area} = \frac{1}{2} \times BC \times d\)
\(203.92 = \frac{1}{2} \times 23.846 \times d\)
\(d = \frac{2 \times 203.92}{23.846} \approx 17.1025 \approx 17.1\text{ km}\) (to 3 s.f.).

**(a)**
M1: For \(\angle BAC = 65^\circ\) seen or used
M1: For substituting correctly into Cosine Rule: \(18^2 + 25^2 - 2(18)(25)\cos(65^\circ)\)
A1: For \(568.6\) or \(569\)
A1: For \(23.8\) or \(23.84\) to \(23.85\)

**(b)**
M1: For using Sine Rule or Cosine Rule to find \(\angle ABC\)
M1: For \(\angle ABC \approx 71.8^\circ\) (or \(71.84^\circ\))
M1: For using back bearing or alternate angles to find South-related angle at \(B\): e.g. \(75^\circ\) or \(255^\circ\)
M1: For \(255 - \text{their } 71.8\) or \(180 + (75 - \text{their } 71.8)\)
A1: For \(183.2\) or \(183\) (accept answers in range \(183\) to \(183.3\))

**(c)**
M1: For finding area of triangle: \(\frac{1}{2} \times 18 \times 25 \times \sin(65^\circ)\) [\(203.9\)]
M1: For equating area to \(\frac{1}{2} \times BC \times d\)
A1: For \(17.1\) or \(17.10\) to \(17.11\)