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Let the total number of balls in the bag be \(5 + x\). The probability of choosing a red ball first is \(\frac{5}{5+x}\). Since the selection is without replacement, the probability of choosing a red ball second is \(\frac{4}{4+x}\). The probability that both balls are red is given by: \(\frac{5}{5+x} \times \frac{4}{4+x} = \frac{5}{18}\). Simplifying this equation: \(\frac{20}{(5+x)(4+x)} = \frac{5}{18}\). Divide both sides by 5: \(\frac{4}{(5+x)(4+x)} = \frac{1}{18}\). Cross-multiplying gives: \((5+x)(4+x) = 72\), which expands to: \(x^2 + 9x + 20 = 72 \Rightarrow x^2 + 9x - 52 = 0\). Factoring the quadratic: \((x - 4)(x + 13) = 0\). Since the number of balls \(x\) must be a positive integer, we have \(x = 4\).

M1: Sets up the probability equation \(\frac{5}{5+x} \times \frac{4}{4+x} = \frac{5}{18}\). M1: Rearranges to form a quadratic equation, e.g., \(x^2 + 9x - 52 = 0\). A1: Correctly solves for \(x = 4\) (rejecting \(x = -13\)).

We can find the coordinates of the minimum point by completing the square for the expression \(x^2 - 10x + 31\). Halving the coefficient of \(x\) gives \(-5\), so we write: \(y = (x - 5)^2 - (-5)^2 + 31\). Simplifying this: \(y = (x - 5)^2 - 25 + 31 \Rightarrow y = (x - 5)^2 + 6\). The minimum value of \((x - 5)^2\) is \(0\), which occurs when \(x = 5\). When \(x = 5\), the minimum value of \(y\) is \(6\). Therefore, the coordinates of the minimum point \(P\) are \((5, 6)\).

M1: Attempts to complete the square, finding \((x - 5)^2\). A1: Correctly completes the square to obtain \((x - 5)^2 + 6\). A1: Correctly identifies coordinates as \((5, 6)\).

To solve \(\frac{3}{x} + \frac{2}{x+2} = 1\), first multiply every term by the common denominator \(x(x+2)\) to clear the fractions: \(3(x+2) + 2x = x(x+2)\). Expand the brackets: \(3x + 6 + 2x = x^2 + 2x\). Simplify and rearrange the terms to form a quadratic equation: \(5x + 6 = x^2 + 2x \Rightarrow x^2 - 3x - 6 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=1, b=-3, c=-6\): \(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-6)}}{2(1)} = \frac{3 \pm \sqrt{9 + 24}}{2} = \frac{3 \pm \sqrt{33}}{2}\). Calculating the values: \(x \approx \frac{3 + 5.7446}{2} \approx 4.37\) and \(x \approx \frac{3 - 5.7446}{2} \approx -1.37\).

M1: Eliminates fractions to get \(3(x+2) + 2x = x(x+2)\) or equivalent. M1: Rearranges into the form \(x^2 - 3x - 6 = 0\). A1: Finds both solutions to 2 decimal places: \(x = 4.37\) and \(x = -1.37\).

The interval from \(t = 0\) to \(t = 3\) is divided into 3 equal strips, so the width of each strip is \(h = 1\) second. We calculate the velocity at each boundary point: For \(t = 0\), \(v_0 = 12(0) - 0^3 = 0\). For \(t = 1\), \(v_1 = 12(1) - 1^3 = 11\). For \(t = 2\), \(v_2 = 12(2) - 2^3 = 16\). For \(t = 3\), \(v_3 = 12(3) - 3^3 = 9\). Using the trapezium rule, the estimated area under the curve (which represents distance) is: \(\text{Area} \approx \frac{1}{2} \times h \times [v_0 + 2(v_1 + v_2) + v_3] = \frac{1}{2} \times 1 \times [0 + 2(11 + 16) + 9] = \frac{1}{2} \times [54 + 9] = 31.5\text{ meters}\).

M1: Calculates the correct boundary values \(v_0 = 0, v_1 = 11, v_2 = 16, v_3 = 9\) (allow one calculation error). M1: Correctly substitutes values into the trapezium rule formula \(\frac{1}{2} \times 1 \times [v_0 + 2(v_1 + v_2) + v_3]\). A1: Obtains 31.5.

The area of a triangle is given by \(\text{Area} = \frac{1}{2} a c \sin B\). Substituting the given values: \(28 = \frac{1}{2} \times 10 \times 7 \sin B \Rightarrow 28 = 35 \sin B\). This gives \(\sin B = \frac{28}{35} = 0.8\). Since angle \(ABC\) (which is angle \(B\)) is obtuse, we find \(B = 180^\circ - \arcsin(0.8) \approx 180^\circ - 53.13^\circ = 126.87^\circ\). Now we use the cosine rule to find the length of \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos B\). Substituting the values: \(AC^2 = 7^2 + 10^2 - 2(7)(10)\cos(126.87^\circ)\). Since \(\cos(126.87^\circ) = -0.6\), we have: \(AC^2 = 49 + 100 - 140(-0.6) = 149 + 84 = 233\). Taking the square root: \(AC = \sqrt{233} \approx 15.264\). To 3 significant figures, the length is \(15.3\text{ cm}\).

M1: Correctly uses the area formula to find \(\sin B = 0.8\) and identifies the obtuse angle \(B \approx 126.87^\circ\) (or \(\cos B = -0.6\)). M1: Uses the cosine rule with their values to calculate \(AC^2\). A1: Correctly evaluates \(AC = 15.3\) (accept 15.26 to 15.3).

First, we find the probability of the union of \(A\) and \(B\): \(P(A \cup B) = 1 - P(A \cup B)' = 1 - 0.15 = 0.85\). Next, we use the addition law of probability: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Substituting the known values: \(0.85 = 0.6 + 0.5 - P(A \cap B) \Rightarrow 0.85 = 1.1 - P(A \cap B) \Rightarrow P(A \cap B) = 0.25\). We want to find \(P(A' \cap B)\), which is the probability of event \(B\) occurring but not event \(A\). This is given by: \(P(A' \cap B) = P(B) - P(A \cap B) = 0.5 - 0.25 = 0.25\).

M1: Calculates \(P(A \cup B) = 0.85\). M1: Uses the addition rule to find \(P(A \cap B) = 0.25\). A1: Correctly evaluates \(P(A' \cap B) = 0.25\).

From the first equation, express \(y\) in terms of \(x\): \(y = 2x + 3\). Substitute this expression into the second equation: \(x^2 + (2x + 3)^2 = 26\). Expand the bracket: \(x^2 + 4x^2 + 12x + 9 = 26\). Simplify and set to zero to form a quadratic equation: \(5x^2 + 12x - 17 = 0\). Factor the quadratic equation: \((5x + 17)(x - 1) = 0\). This gives two solutions: \(x = -\frac{17}{5}\) or \(x = 1\). Since we require the positive value of \(x\), we choose \(x = 1\). Substitute \(x = 1\) back into the linear equation to find \(y\): \(y = 2(1) + 3 = 5\). Thus, the solution is \(x = 1, y = 5\).

M1: Substitutes \(y = 2x+3\) into the quadratic equation. M1: Correctly expands and forms the quadratic equation \(5x^2 + 12x - 17 = 0\). A1: Finds the correct positive value \(x = 1\) and corresponding \(y = 5\).

Substitute the equation of the line \(y = 2x\) into the equation of the circle \(x^2 + y^2 = 50\): \(x^2 + (2x)^2 = 50 \Rightarrow x^2 + 4x^2 = 50 \Rightarrow 5x^2 = 50\). Dividing by 5 gives: \(x^2 = 10 \Rightarrow x = \pm\sqrt{10}\). Now substitute these values back into \(y = 2x\) to find the corresponding \(y\)-coordinates: When \(x = \sqrt{10}\), \(y = 2\sqrt{10}\). When \(x = -\sqrt{10}\), \(y = -2\sqrt{10}\). Therefore, the coordinates of the two intersection points are \((\sqrt{10}, 2\sqrt{10})\) and \((-\sqrt{10}, -2\sqrt{10})\).

M1: Substitutes \(y = 2x\) into the circle equation and simplifies to \(5x^2 = 50\) or equivalent. M1: Correctly solves for \(x\) to get \(x = \pm\sqrt{10}\). A1: Correctly states both coordinates in exact form: \((\sqrt{10}, 2\sqrt{10})\) and \((-\sqrt{10}, -2\sqrt{10})\).

The probability of choosing a red counter first is \(\frac{x}{x+3}\). Since the counter is not replaced, the probability of choosing a second red counter is \(\frac{x-1}{x+2}\). The probability that both are red is given by: \(\frac{x}{x+3} \times \frac{x-1}{x+2} = \frac{5}{12}\). Multiplying both sides by \(12(x+3)(x+2)\) gives: \(12x(x-1) = 5(x+3)(x+2)\). Expanding both sides: \(12x^2 - 12x = 5(x^2 + 5x + 6)\), which simplifies to \(12x^2 - 12x = 5x^2 + 25x + 30\). Rearranging to form a quadratic equation: \(7x^2 - 37x - 30 = 0\). Factoring this quadratic gives: \((7x+5)(x-6) = 0\). Since \(x\) must be a positive integer, we reject the negative solution \(x = -\frac{5}{7}\). Thus, \(x = 6\).

M1 for setting up the initial probability equation: \(\frac{x}{x+3} \times \frac{x-1}{x+2} = \frac{5}{12}\). M1 for simplifying to a correct quadratic equation, e.g., \(7x^2 - 37x - 30 = 0\). A1 for \(x = 6\) (with the negative solution rejected or omitted).

Equating the equations of the line and the curve: \(2x + k = x^2 - 4x + 14\). Rearranging this into a standard quadratic form gives: \(x^2 - 6x + (14 - k) = 0\). For the line to be a tangent to the curve, the quadratic equation must have exactly one real root, meaning the discriminant must be zero: \(b^2 - 4ac = 0\). Substituting \(a = 1\), \(b = -6\), and \(c = 14 - k\) into the discriminant formula: \((-6)^2 - 4(1)(14 - k) = 0\). This simplifies to: \(36 - 56 + 4k = 0\), which gives \(-20 + 4k = 0\). Solving for \(k\) yields \(4k = 20\), so \(k = 5\).

M1 for equating the line and the curve and rearranging to form a quadratic equation, e.g., \(x^2 - 6x + 14 - k = 0\). M1 for setting the discriminant equal to zero: \((-6)^2 - 4(1)(14-k) = 0\). A1 for \(k = 5\).

To solve the equation, multiply every term by the common denominator \((x + 1)(x - 2)\): \(5(x - 2) + 2(x + 1) = 2(x + 1)(x - 2)\). Expand both sides: \(5x - 10 + 2x + 2 = 2(x^2 - x - 2)\). Simplify the left side and expand the right side: \(7x - 8 = 2x^2 - 2x - 4\). Rearrange all terms to one side to set the equation to zero: \(2x^2 - 9x + 4 = 0\). Factorise the quadratic equation: \((2x - 1)(x - 4) = 0\). Solving for \(x\) gives \(2x - 1 = 0\) or \(x - 4 = 0\), which results in \(x = 0.5\) or \(x = 4\).

M1 for multiplying through by the common denominator to obtain a linear or quadratic equation: \(5(x-2) + 2(x+1) = 2(x+1)(x-2)\). M1 for simplifying to a standard quadratic equation: \(2x^2 - 9x + 4 = 0\). A1 for both solutions: \(x = 0.5\) (or \(\frac{1}{2}\)) and \(x = 4\).

Using the formula for the area of a triangle, \(\text{Area} = \frac{1}{2} b c \sin(A)\), we have: \(28 = \frac{1}{2} \times 7.2 \times 9.5 \times \sin(BAC)\). Simplifying this gives: \(28 = 34.2 \sin(BAC)\), so \(\sin(BAC) = \frac{28}{34.2} \approx 0.8187\). Since angle \(BAC\) is obtuse, we find the angle in the second quadrant: \(BAC = 180^\circ - \sin^{-1}(0.8187) \approx 180^\circ - 54.95^\circ = 125.05^\circ\). Now, apply the Cosine rule to find the length of \(BC\): \(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(BAC)\). Substituting the known values: \(BC^2 = 7.2^2 + 9.5^2 - 2(7.2)(9.5)\cos(125.05^\circ)\). This gives: \(BC^2 = 51.84 + 90.25 - 136.8 \times (-0.5743) \approx 142.09 + 78.56 = 220.65\). Taking the square root: \(BC = \sqrt{220.65} \approx 14.854\text{ cm}\). To 3 significant figures, the length of \(BC\) is \(14.9\text{ cm}\).

M1 for using the area formula to find the obtuse angle: \(BAC = 180^\circ - \sin^{-1}\left(\frac{28}{34.2}\right) \approx 125.1^\circ\) (accept \(125^\circ\) or \(125.05^\circ\)). M1 for using the Cosine rule with their obtuse angle: \(BC^2 = 7.2^2 + 9.5^2 - 2(7.2)(9.5)\cos(125.1^\circ)\). A1 for \(14.9\) (accept answers in the range \(14.8\) to \(14.9\)).

The total number of counters in the bag is \(4 + 3 + n = n + 7\).

Since two counters are drawn without replacement, we calculate the individual probabilities for each color:
- Probability of 2 Red: \(P(\text{Red, Red}) = \frac{4}{n+7} \times \frac{3}{n+6} = \frac{12}{(n+7)(n+6)}\)
- Probability of 2 Blue: \(P(\text{Blue, Blue}) = \frac{3}{n+7} \times \frac{2}{n+6} = \frac{6}{(n+7)(n+6)}\)
- Probability of 2 Yellow: \(P(\text{Yellow, Yellow}) = \frac{n}{n+7} \times \frac{n-1}{n+6} = \frac{n(n-1)}{(n+7)(n+6)}\)

The probability of drawing two counters of the same colour is:
\(P(\text{Same Colour}) = \frac{12 + 6 + n(n-1)}{(n+7)(n+6)} = \frac{5}{18}\)

Simplifying the fraction:
\(\frac{n^2 - n + 18}{n^2 + 13n + 42} = \frac{5}{18}\)

Cross-multiply to solve:
\(18(n^2 - n + 18) = 5(n^2 + 13n + 42)\)
\(18n^2 - 18n + 324 = 5n^2 + 65n + 210\)
\(13n^2 - 83n + 114 = 0\)

Factorising this quadratic equation:
\((n - 2)(13n - 57) = 0\)

This gives \(n = 2\) or \(n = \frac{57}{13}\).
Since \(n\) must be a positive integer, we have \(n = 2\).

M1: Writes down expressions for individual probabilities or attempts to sum them: \(\frac{4 \times 3}{(n+7)(n+6)} + \frac{3 \times 2}{(n+7)(n+6)} + \frac{n(n-1)}{(n+7)(n+6)}\).
M1: Equates sum to \(\frac{5}{18}\) and attempts to clear fractions, arriving at a quadratic expression.
A1: Correctly derives the simplified quadratic equation \(13n^2 - 83n + 114 = 0\) (or equivalent).
A1.5: Factorises or solves to find \(n = 2\) and rejects the non-integer solution.

Substitute the given coordinates into the equation \(y = a b^x\):
1) For \((1, 15)\):
\(15 = a b^1 \implies a b = 15\)

2) For \((3, 135)\):
\(135 = a b^3\)

We can divide the second equation by the first equation to eliminate \(a\):
\(\frac{a b^3}{a b} = \frac{135}{15}\)
\(b^2 = 9\)

Since \(b\) is a positive constant:
\(b = 3\)

Substitute \(b = 3\) back into the first equation:
\(a(3) = 15 \implies a = 5\)

So, the values are \(a = 5\) and \(b = 3\).

M1: Sets up both simultaneous equations: \(a b = 15\) and \(a b^3 = 135\).
M1: Eliminates \(a\) by dividing the equations to get \(b^2 = 9\) (or substitution equivalent).
A1: Finds \(b = 3\) (accepting \(b = 3\) with justification that \(b > 0\)).
A1.5: Finds \(a = 5\).

Multiply the entire equation by the common denominator \((x - 1)(x + 2)\):
\[ 2(x + 2) + 3(x - 1) = 2(x - 1)(x + 2) \]

Expand both sides of the equation:
\[ 2x + 4 + 3x - 3 = 2(x^2 + x - 2) \]
\[ 5x + 1 = 2x^2 + 2x - 4 \]

Rearrange the equation to form a quadratic equal to zero:
\[ 2x^2 - 3x - 5 = 0 \]

Factorise the quadratic expression:
\[ (2x - 5)(x + 1) = 0 \]

This gives the solutions:
\[ 2x - 5 = 0 \implies x = 2.5 \]
\[ x + 1 = 0 \implies x = -1 \]

M1: Multiplies through by the common denominator \((x-1)(x+2)\) to remove fractions.
M1: Correctly expands both sides to get \(5x + 1\) and \(2x^2 + 2x - 4\) (or equivalent).
A1: Rearranges to obtain the correct quadratic equation \(2x^2 - 3x - 5 = 0\).
A1.5: Solves the quadratic to find both correct solutions: \(x = 2.5\) and \(x = -1\).

The total distance is estimated by calculating the area under the velocity-time curve using the trapezium rule. The width of each of the 4 intervals is \(h = 5\).

The area of each trapezium is calculated as follows:
- First trapezium (from \(t = 0\) to \(t = 5\)):
\[ \text{Area}_1 = \frac{0 + 4}{2} \times 5 = 10\text{ m} \]
- Second trapezium (from \(t = 5\) to \(t = 10\)):
\[ \text{Area}_2 = \frac{4 + 10}{2} \times 5 = 35\text{ m} \]
- Third trapezium (from \(t = 10\) to \(t = 15\)):
\[ \text{Area}_3 = \frac{10 + 18}{2} \times 5 = 70\text{ m} \]
- Fourth trapezium (from \(t = 15\) to \(t = 20\)):
\[ \text{Area}_4 = \frac{18 + 25}{2} \times 5 = 107.5\text{ m} \]

Adding these areas together gives the total estimated distance:
\[ \text{Total Distance} = 10 + 35 + 70 + 107.5 = 222.5\text{ m} \]

M1: Identifies the width of each interval as \(h = 5\).
M1: Attempts to use the trapezium area formula for at least two intervals.
A1: Correctly calculates the areas of all four trapeziums: 10, 35, 70, and 107.5.
A1.5: Sums the areas correctly to get the final answer of 222.5.

First, sketch the journey to determine the interior angle \(\angle ABC\) of triangle ABC:
- The bearing of B from A is \(060^\circ\), which means the line BA makes an angle of \(180^\circ - 060^\circ = 120^\circ\) with North (counter-clockwise).
- The bearing of C from B is \(140^\circ\) (clockwise from North).
- Therefore, the interior angle at B is:
\[ \angle ABC = 360^\circ - 120^\circ - 140^\circ = 100^\circ \]

Using the Cosine Rule to find side \(AC\):
\[ AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos(\angle ABC) \]
\[ AC^2 = 15^2 + 22^2 - 2(15)(22) \cos(100^\circ) \]
\[ AC^2 = 225 + 484 - 660 \cos(100^\circ) \]
\[ AC^2 \approx 709 - 660(-0.173648) \]
\[ AC^2 \approx 709 + 114.608 = 823.608 \]
\[ AC = \sqrt{823.608} \approx 28.6986\text{ km} \]

To 3 significant figures, the distance is \(28.7\text{ km}\).

M1: Attempts to find the interior angle \(\angle ABC\) using bearings properties.
A1: Correctly finds \(\angle ABC = 100^\circ\).
M1: Correct substitution of values into the Cosine Rule: \(15^2 + 22^2 - 2(15)(22)\cos(100^\circ)\).
A1.5: Correctly evaluates to find the final distance as 28.7 km (to 3 sig figs).

Substitute the expression for \(y\) from the equation of the line into the equation of the circle:
\[ x^2 + (2x - 5)^2 = 25 \]

Expand the brackets:
\[ x^2 + 4x^2 - 20x + 25 = 25 \]
\[ 5x^2 - 20x = 0 \]

Factorise the quadratic equation:
\[ 5x(x - 4) = 0 \]

This gives two values for \(x\):
\[ x = 0 \quad \text{or} \quad x = 4 \]

Now, substitute these \(x\)-values back into the line equation \(y = 2x - 5\) to find the corresponding \(y\)-coordinates:
- For \(x = 0\):
\[ y = 2(0) - 5 = -5 \implies (0, -5) \]
- For \(x = 4\):
\[ y = 2(4) - 5 = 3 \implies (4, 3) \]

The points of intersection are \((0, -5)\) and \((4, 3)\).

M1: Substitutes \(y = 2x - 5\) into the circle equation.
M1: Expands and simplifies to a quadratic equation, e.g., \(5x^2 - 20x = 0\).
A1: Solves the quadratic to find both \(x = 0\) and \(x = 4\).
A1.5: Determines both correct coordinates, \((0, -5)\) and \((4, 3)\).

Let \(r\) be the number of Red balls, \(g\) be the number of Green balls, and \(b\) be the number of Blue balls. We are given that \(r + g + b = 30\). First, consider the probability of selecting two Red balls without replacement: \(P(\text{Red, Red}) = \frac{r}{30} \times \frac{r-1}{29} = \frac{2}{29}\). Multiplying both sides by 29 gives: \(\frac{r(r-1)}{30} = 2 \implies r(r-1) = 60 \implies r^2 - r - 60 = 0\). Factorising this quadratic equation gives \((r-10)(r+9) = 0\). Since the number of balls must be positive, we have \(r = 10\). Thus, the sum of Green and Blue balls is: \(g + b = 30 - 10 = 20\). Next, consider the probability of drawing three balls of different colours without replacement. The three balls can be drawn in \(3! = 6\) different orders (RGB, RBG, GRB, GBR, BRG, BGR). The probability of any one of these specific orders (for example, RGB) is: \(P(RGB) = \frac{r}{30} \times \frac{g}{29} \times \frac{b}{28}\). Since all 6 orders have the same probability, the total probability that all three balls are of different colours is: \(P(\text{all different}) = 6 \times \frac{10 \times g \times b}{30 \times 29 \times 28} = \frac{48}{203}\). Simplifying the left-hand side: \(\frac{60gb}{24360} = \frac{gb}{406}\). This gives: \(\frac{gb}{406} = \frac{48}{203} \implies gb = \frac{48 \times 406}{203} = 48 \times 2 = 96\). We now have a system of equations: 1) \(g + b = 20\) and 2) \(gb = 96\). Substituting \(b = 20 - g\) into the second equation: \(g(20 - g) = 96 \implies g^2 - 20g + 96 = 0\). Factorising this quadratic equation gives \((g-12)(g-8) = 0\). Thus, \(g = 12\) or \(g = 8\). Since we are given that there are more Green balls than Blue balls (\(g > b\)), we must have \(g = 12\) and \(b = 8\). Therefore, the number of balls of each colour is Red: 10, Green: 12, Blue: 8.

M1: Sets up the probability equation for two Red balls: \(\frac{r(r-1)}{30 \times 29} = \frac{2}{29}\).
A1: Correctly simplifies and solves the quadratic to find \(r = 10\).
M1: States or uses that \(g + b = 20\).
M1: Recognises that there are \(3! = 6\) ways to arrange three different-coloured balls.
M1: Expresses the probability of drawing one of each colour as \(6 \times \frac{10 \times g \times b}{30 \times 29 \times 28}\).
A1: Equates this to \(\frac{48}{203}\) and simplifies to find \(gb = 96\).
M1: Formulates a quadratic equation to solve for \(g\) and \(b\), e.g., \(g^2 - 20g + 96 = 0\).
A1: Solves the quadratic to find the values 12 and 8.
A1: Uses the condition \(g > b\) to state the final correct counts: Red: 10, Green: 12, Blue: 8.

A cuboid has 12 edges: 4 of length \(x\), 4 of length \(x+2\), and 4 of length \(y\). The sum of the edge lengths is: \(4x + 4(x+2) + 4y = 88 \implies x + (x+2) + y = 22 \implies 2x + 2 + y = 22 \implies y = 20 - 2x\) [Equation 1]. The total surface area of the cuboid is: \(A = 2(x(x+2) + xy + (x+2)y) = 288 \implies x^2 + 2x + xy + xy + 2y = 144 \implies x^2 + 2x + 2xy + 2y = 144\) [Equation 2]. Substituting Equation 1 into Equation 2: \(x^2 + 2x + 2x(20 - 2x) + 2(20 - 2x) = 144 \implies x^2 + 2x + 40x - 4x^2 + 40 - 4x = 144 \implies -3x^2 + 38x + 40 = 144\). Rearranging this gives: \(3x^2 - 38x + 104 = 0\). Solving this quadratic using factorisation: \((3x - 26)(x - 4) = 0\). This gives two possible values for \(x\): \(x = 4\) or \(x = \frac{26}{3}\). Case 1: If \(x = 4\), the dimensions are \(4\text{ cm}\), \(6\text{ cm}\), and \(y = 20 - 2(4) = 12\text{ cm}\). The volume is \(V_1 = 4 \times 6 \times 12 = 288\text{ cm}^3\). Case 2: If \(x = \frac{26}{3}\), the dimensions are \(\frac{26}{3}\text{ cm}\), \(\frac{32}{3}\text{ cm}\), and \(y = 20 - 2(\frac{26}{3}) = \frac{8}{3}\text{ cm}\). The volume is \(V_2 = \frac{26}{3} \times \frac{32}{3} \times \frac{8}{3} = \frac{6656}{27} \approx 246.5185\text{ cm}^3\). Giving the answers to 1 decimal place, the volumes are \(288\text{ cm}^3\) (or \(288.0\text{ cm}^3\)) and \(246.5\text{ cm}^3\).

M1: Writes correct equation for total edge lengths: \(4x + 4(x+2) + 4y = 88\).
A1: Simplifies to get a correct linear expression for \(y\) in terms of \(x\), e.g., \(y = 20 - 2x\).
M1: Writes correct equation for total surface area: \(2(x(x+2) + y(x+2) + xy) = 288\).
A1: Simplifies to \(x^2 + 2x + 2xy + 2y = 144\).
M1: Substitutes \(y = 20 - 2x\) into the surface area equation.
A1: Correctly derives the quadratic equation \(3x^2 - 38x + 104 = 0\) (or equivalent).
M1: Solves the quadratic to find \(x = 4\) and \(x = \frac{26}{3}\).
M1: Evaluates the dimensions and calculates the volumes for both cases.
A1: Correctly states the two volumes as \(288\text{ cm}^3\) (or \(288.0\)) and \(246.5\text{ cm}^3\) to 1 decimal place.

Let \(O\) be the point on the horizontal plane directly below \(P\) and \(T\), so \(T, P, O\) form a vertical straight line with \(\angle TOB = \angle POB = 90^\circ\). In triangle \(ABT\): \(\angle TAB = 20^\circ\). Since \(\angle TBO = 34^\circ\), the interior angle is \(\angle TBA = 180^\circ - 34^\circ = 146^\circ\). Thus, the angle \(\angle ATB = 180^\circ - (20^\circ + 146^\circ) = 14^\circ\). The length of \(AB = 150\text{ m}\). Applying the Sine Rule in triangle \(ABT\): \(\frac{BT}{\sin(20^\circ)} = \frac{150}{\sin(14^\circ)} \implies BT = \frac{150 \times \sin(20^\circ)}{\sin(14^\circ)}\). Using a calculator: \(BT \approx \frac{150 \times 0.342020}{0.241922} \approx 212.064\text{ m}\). Now, consider the right-angled triangle \(TBO\): \(TO = BT \times \sin(34^\circ) \approx 212.064 \times \sin(34^\circ) \approx 118.579\text{ m}\) and \(BO = BT \times \cos(34^\circ) \approx 212.064 \times \cos(34^\circ) \approx 175.811\text{ m}\). In the right-angled triangle \(PBO\): The angle of elevation of \(P\) from \(B\) is \(28^\circ\). So \(\tan(28^\circ) = \frac{PO}{BO} \implies PO = BO \times \tan(28^\circ) \approx 175.811 \times \tan(28^\circ) \approx 93.480\text{ m}\). Thus, the height of the hill is \(PO \approx 93.5\text{ m}\). The height of the radio mast is \(h = TP = TO - PO \approx 118.579 - 93.480 = 25.099\text{ m}\). Thus, the height of the radio mast is \(h \approx 25.1\text{ m}\).

M1: Formulates the interior angles of triangle \(ABT\), finding \(\angle TBA = 146^\circ\) or equivalent.
A1: Calculates \(\angle ATB = 14^\circ\) correctly.
M1: Correctly applies the Sine Rule to triangle \(ABT\): \(\frac{BT}{\sin(20^\circ)} = \frac{150}{\sin(14^\circ)}\).
A1: Calculates \(BT \approx 212\text{ m}\) (or \(AT \approx 347\text{ m}\)) with work shown.
M1: Uses right-angled trigonometry on triangle \(TBO\) to find \(TO = BT \sin(34^\circ) \approx 118.6\text{ m}\).
M1: Uses right-angled trigonometry on triangle \(TBO\) to find \(BO = BT \cos(34^\circ) \approx 175.8\text{ m}\).
M1: Uses right-angled trigonometry on triangle \(PBO\) to find \(PO = BO \tan(28^\circ)\).
A1: Finds the height of the hill \(PO \approx 93.5\text{ m}\) (accept range 93.3 - 93.7).
A1: Finds the height of the mast \(h \approx 25.1\text{ m}\) (accept range 24.9 - 25.3).

First, find the equation of the tangent line \(L\) at \(P(3, 4)\). The center of the circle is \(O(0, 0)\). The gradient of the radius \(OP\) is \(m = \frac{4 - 0}{3 - 0} = \frac{4}{3}\). Since the tangent is perpendicular to the radius, the gradient of \(L\) is \(m_L = -\frac{3}{4}\). Using \(P(3, 4)\), the equation of \(L\) is: \(y - 4 = -\frac{3}{4}(x - 3) \implies y = -\frac{3}{4}x + \frac{25}{4}\). This can be written as \(3x + 4y = 25\). Second, find the equation of the line \(M\). Since \(M\) is perpendicular to \(L\), its gradient is \(m_M = \frac{4}{3}\). Since \(M\) passes through \(Q(0, -5)\), its \(y\)-intercept is \(-5\). Thus, the equation of \(M\) is \(y = \frac{4}{3}x - 5\). Third, find the intersection point of \(L\) and \(M\) by setting their equations equal: \(-\frac{3}{4}x + \frac{25}{4} = \frac{4}{3}x - 5\). Multiplying by 12 gives: \(-9x + 75 = 16x - 60 \implies 25x = 135 \implies x = 5.4\). Substituting \(x = 5.4\) into \(M\): \(y = \frac{4}{3}(5.4) - 5 = 7.2 - 5 = 2.2\). So the intersection point is \(R(5.4, 2.2)\). Fourth, the triangle is bounded by \(L\), \(M\), and the \(y\)-axis. The vertices are the intersection point \(R(5.4, 2.2)\), the \(y\)-intercept of \(L\) at \(B(0, 6.25)\), and the \(y\)-intercept of \(M\) at \(Q(0, -5)\). The base of the triangle along the \(y\)-axis is the distance between \(B\) and \(Q\): \(b = 6.25 - (-5) = 11.25\). The perpendicular height of the triangle is the \(x\)-coordinate of the intersection point \(R\), which is \(h = 5.4\). The area of the triangle is: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 11.25 \times 5.4 = 30.375\).

M1: Calculates the gradient of the radius \(OP\) as \(\frac{4}{3}\).
M1: Uses the perpendicular gradient rule to find the gradient of \(L\) as \(-\frac{3}{4}\).
A1: Writes the correct equation for \(L\), e.g., \(y = -0.75x + 6.25\) or \(3x + 4y = 25\).
M1: Finds the gradient of \(M\) as \(\frac{4}{3}\) and writes the equation of \(M\), e.g., \(y = \frac{4}{3}x - 5\).
M1: Equates the equations of \(L\) and \(M\) to find their intersection.
A1: Finds the correct coordinates of intersection: \((5.4, 2.2)\).
M1: Identifies the two \(y\)-intercepts as \((0, 6.25)\) and \((0, -5)\).
M1: Uses the formula for the area of a triangle, \(\frac{1}{2} \times b \times h\), with base \(11.25\) and height \(5.4\).
A1: Calculates the exact area of the triangle as \(30.375\) (or \(\frac{243}{8}\)).

Let the total number of counters be \(5 + x\). The probability of choosing a red counter on the first draw is \(\frac{5}{5+x}\). Since the counter is not replaced, the probability of choosing a red counter on the second draw is \(\frac{4}{4+x}\). The probability that both are red is:

\(\frac{5}{5+x} \times \frac{4}{4+x} = \frac{2}{9}\)

This simplifies to:

\(\frac{20}{x^2 + 9x + 20} = \frac{2}{9}\)

\(180 = 2(x^2 + 9x + 20)\)

\(90 = x^2 + 9x + 20\)

\(x^2 + 9x - 70 = 0\)

Factorising the quadratic equation gives:

\((x + 14)(x - 5) = 0\)

Since \(x\) must be a positive integer, we have \(x = 5\).

M1: For a correct probability equation for two red counters, e.g., \(\frac{5}{5+x} \times \frac{4}{4+x} = \frac{2}{9}\)
M1: For simplifying the equation to a standard quadratic form, e.g., \(x^2 + 9x - 70 = 0\)
A1: For \(5\) (with the negative solution \(x = -14\) discarded)

The turning point of the quadratic curve \(y = x^2 + kx + 15\) occurs where \(x = -\frac{k}{2}\). Since the turning point is at \((3, p)\), we have:

\(-\frac{k}{2} = 3 \implies k = -6\)

Thus, the equation of the curve is \(y = x^2 - 6x + 15\).

To find the \(y\)-coordinate \(p\) of the turning point, substitute \(x = 3\) into the equation:

\(p = 3^2 - 6(3) + 15 = 9 - 18 + 15 = 6\).

Alternatively, by completing the square:

\(y = (x - 3)^2 + p = x^2 - 6x + 9 + p\).

Comparing this to the original equation gives:

\(9 + p = 15 \implies p = 6\).

M1: For identifying \(k = -6\) (e.g., using \(x = -\frac{k}{2} = 3\) or by completing the square to get \((x - 3)^2\))
M1: For substituting \(x = 3\) and \(k = -6\) into the original equation, or setting \(9 + p = 15\)
A1: For \(6\)

From the linear equation, rearrange to express \(y\) in terms of \(x\):

\(y = 2x - 5\)

Substitute this into the quadratic equation:

\(x^2 + (2x - 5)^2 = 25\)

\(x^2 + 4x^2 - 20x + 25 = 25\)

\(5x^2 - 20x = 0\)

Factorise the equation:

\(5x(x - 4) = 0\)

This gives two solutions: \(x = 0\) or \(x = 4\).

Since we are given that \(x > 0\), we choose \(x = 4\).

Substitute \(x = 4\) back into the rearranged linear equation to find \(y\):

\(y = 2(4) - 5 = 3\).

Therefore, the solution is \((4, 3)\).

M1: For substituting \(y = 2x - 5\) (or equivalent) into the quadratic equation to form an equation in one variable
M1: For simplifying to \(5x^2 - 20x = 0\) and solving to find \(x = 4\)
A1: For \((4, 3)\)

The distance travelled is represented by the area under the velocity-time graph, which is a trapezium with parallel sides of \(6\) seconds (from \(t = 4\) to \(t = 10\)) and \(12\) seconds (from \(t = 0\) to \(t = 12\)), and a height of \(V\).

Area of a trapezium = \(\frac{1}{2}(a + b)h = \frac{1}{2}(6 + 12)V = 9V\).

We are given that the total distance is \(180\) metres, so:

\(9V = 180 \implies V = 20\).

M1: For expressing the distance in terms of \(V\), e.g., \(\frac{1}{2} \times (6 + 12) \times V\) or \(\frac{1}{2}(4)(V) + 6V + \frac{1}{2}(2)(V)\)
M1: For setting up the equation \(9V = 180\) (or equivalent)
A1: For \(20\)

Using the Cosine Rule:

\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)

Substitute the given values:

\(AC^2 = 6^2 + (5\sqrt{2})^2 - 2(6)(5\sqrt{2})\cos(45^\circ)\)

Since \(\cos(45^\circ) = \frac{1}{\sqrt{2}}\), we have:

\(AC^2 = 36 + 50 - 60\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\)

\(AC^2 = 86 - 60\)

\(AC^2 = 26\)

Taking the square root gives the exact length:

\(AC = \sqrt{26}\).

M1: For correct substitution into the Cosine Rule: \(AC^2 = 6^2 + (5\sqrt{2})^2 - 2(6)(5\sqrt{2})\cos(45^\circ)\)
M1: For evaluating \(\cos(45^\circ) = \frac{1}{\sqrt{2}}\) (or \(\frac{\sqrt{2}}{2}\)) and simplifying to find \(AC^2 = 26\)
A1: For \(\sqrt{26}\)

First, find the probability that Mia loses the game:

\(P(\text{Lose}) = 1 - 0.4 - 0.3 = 0.3\).

Next, find the probability of NOT playing again for each outcome:
- If she wins, the probability of not playing again is \(1 - 0.3 = 0.7\).
- If she draws, the probability of not playing again is \(1 - 0.5 = 0.5\).
- If she loses, the probability of not playing again is \(1 - 0.2 = 0.8\).

Use the law of total probability to find the overall probability of not playing again:

\(P(\text{Not play again}) = (0.4 \times 0.7) + (0.3 \times 0.5) + (0.3 \times 0.8)\)

\(P(\text{Not play again}) = 0.28 + 0.15 + 0.24 = 0.67\).

M1: For finding the probability of losing, \(1 - 0.4 - 0.3 = 0.3\)
M1: For a correct probability sum, e.g., \((0.4 \times 0.7) + (0.3 \times 0.5) + (0.3 \times 0.8)\)
A1: For \(0.67\) (or \(\frac{67}{100}\))

Multiply the entire equation by the common denominator \((x+2)(x-1)\):

\(7(x-1) + 4(x+2) = 2(x+2)(x-1)\)

Expand both sides:

\(7x - 7 + 4x + 8 = 2(x^2 + x - 2)\)

\(11x + 1 = 2x^2 + 2x - 4\)

Rearrange the terms to form a quadratic equation set to zero:

\(2x^2 - 9x - 5 = 0\)

Factorise the quadratic equation:

\((2x + 1)(x - 5) = 0\)

Thus, the solutions are \(x = 5\) or \(x = -0.5\).

M1: For multiplying by the common denominator to get \(7(x-1) + 4(x+2) = 2(x+2)(x-1)\)
M1: For expanding and simplifying to a standard quadratic form \(2x^2 - 9x - 5 = 0\)
A1: For both solutions: \(x = 5\) and \(x = -0.5\) (or \(x = -\frac{1}{2}\))

At the points of intersection, the \(y\)-values are equal, so we set the two equations equal to each other:

\(2x + 5 = x^2 + x - 1\)

Rearrange this equation into standard quadratic form:

\(x^2 - x - 6 = 0\)

Factorise the quadratic equation:

\((x - 3)(x + 2) = 0\)

This gives the \(x\)-coordinates of the intersection points:

\(x = 3\) or \(x = -2\).

Substitute these \(x\)-values back into the linear equation \(y = 2x + 5\) to find the corresponding \(y\)-coordinates:
- For \(x = 3\): \(y = 2(3) + 5 = 11\).
- For \(x = -2\): \(y = 2(-2) + 5 = 1\).

Therefore, the coordinates of the points of intersection are \((3, 11)\) and \((-2, 1)\).

M1: For equating the linear and quadratic expressions: \(2x + 5 = x^2 + x - 1\)
M1: For solving the quadratic equation to obtain \(x = 3\) and \(x = -2\)
A1: For both points \((3, 11)\) and \((-2, 1)\) correct

The probability of choosing a yellow counter first is \(\frac{3}{n}\).

Since the counter is not replaced, there are now \(n-1\) counters left, of which 2 are yellow.

The probability of choosing a second yellow counter is \(\frac{2}{n-1}\).

The probability that both are yellow is:
\[\frac{3}{n} \times \frac{2}{n-1} = \frac{6}{n(n-1)}\]

We are given that this probability is \(\frac{1}{15}\), so:
\[\frac{6}{n(n-1)} = \frac{1}{15}\]

Cross-multiplying gives:
\[6 \times 15 = n(n-1)\]
\[90 = n^2 - n\]
\[n^2 - n - 90 = 0\]

Factorising the quadratic equation:
\[(n - 10)(n + 9) = 0\]

This gives \(n = 10\) or \(n = -9\).

Since the number of counters \(n\) must be positive, we have \(n = 10\).

M1 for setting up the equation: \(\frac{3}{n} \times \frac{2}{n-1} = \frac{1}{15}\) (or equivalent).
M1 for establishing the quadratic equation \(n^2 - n - 90 = 0\) and attempting to solve by factorisation to find \((n-10)(n+9) = 0\) (or equivalent).
A1 for \(n = 10\) (accept 10, reject -9).

Method 1: Completing the square

A quadratic curve with turning point \((p, q)\) can be written in the form:
\[y = (x - p)^2 + q\]

Given the turning point is \((4, -3)\), we substitute \(p = 4\) and \(q = -3\):
\[y = (x - 4)^2 - 3\]

Expanding the bracket:
\[y = (x^2 - 8x + 16) - 3\]
\[y = x^2 - 8x + 13\]

Comparing this with \(y = x^2 + bx + c\):
\[b = -8\]
\[c = 13\]

Method 2: Using the axis of symmetry

The \(x\)-coordinate of the turning point of \(y = ax^2 + bx + c\) is given by \(x = -\frac{b}{2a}\).
Here, \(a = 1\) and the \(x\)-coordinate is \(4\):
\[4 = -\frac{b}{2(1)} \implies b = -8\]

Substitute \(x = 4\), \(y = -3\), and \(b = -8\) into the equation of the curve:
\[-3 = (4)^2 + (-8)(4) + c\]
\[-3 = 16 - 32 + c\]
\[-3 = -16 + c\]
\[c = 13\]

M1 for writing the equation in completed square form \((x - 4)^2 - 3\) OR using the formula for the axis of symmetry to find \(-\frac{b}{2} = 4\).
M1 for expanding to \(x^2 - 8x + 13\) OR substituting \(x = 4\), \(y = -3\), and their \(b\) into the equation to solve for \(c\).
A1 for \(b = -8\) and \(c = 13\).

Let the total number of counters in the bag be \(n + 5\). The probability of choosing a red counter first is \(\frac{5}{n + 5}\). Since the first counter is not replaced, there are now 4 red counters left, and the total number of counters is \(n + 4\). The probability of choosing a red counter second is \(\frac{4}{n + 4}\). The probability of choosing two red counters is \(\frac{5}{n + 5} \times \frac{4}{n + 4} = \frac{20}{(n+5)(n+4)}\). We are given that this probability is \(\frac{2}{9}\), so \(\frac{20}{(n+5)(n+4)} = \frac{2}{9}\). Cross-multiplying gives \(180 = 2(n+5)(n+4)\), which simplifies to \(90 = (n+5)(n+4)\). Expanding this, we get \(90 = n^2 + 9n + 20\), which rearranges to the quadratic equation \(n^2 + 9n - 70 = 0\). Factorising the quadratic equation gives \((n + 14)(n - 5) = 0\). Since the number of blue counters \(n\) must be a positive integer, we discard the negative solution \(n = -14\). Therefore, the value of \(n\) is 5.

M1: Sets up the probability equation \(\frac{5}{n+5} \times \frac{4}{n+4} = \frac{2}{9}\). M1: Expands and rearranges to form a quadratic equation, e.g., \(n^2 + 9n - 70 = 0\). M1: Factorises the quadratic equation to \((n+14)(n-5) = 0\). A1: Finds the roots \(n = 5\) and \(n = -14\). A1: Rejects the negative root to conclude that \(n = 5\).

Let \(x\) be the number of students who study all three languages: Spanish, French, and German. Using the principle of inclusion-exclusion for the total number of students: \(|S \cup F \cup G| = |S| + |F| + |G| - |S \cap F| - |F \cap G| - |S \cap G| + |S \cap F \cap G|\). Substituting the given values: \(80 = 35 + 40 + 33 - 12 - 9 - 14 + x\), which simplifies to \(80 = 73 + x\), giving \(x = 7\). So, 7 students study all three languages. We want to find the conditional probability that a student studies Spanish but not German, given that they study French. The total number of students studying French is 40. Among these 40 French students, those who study Spanish but not German are those studying both French and Spanish minus those studying all three languages: \(12 - 7 = 5\) students. Therefore, the required probability is \(\frac{5}{40} = \frac{1}{8}\).

M1: Applies the principle of inclusion-exclusion or draws an equivalent Venn diagram. M1: Correctly calculates that 7 students study all three languages. M1: Identifies that the number of students who study French and Spanish but not German is \(12 - 7 = 5\). M1: Sets up the conditional probability fraction with a denominator of 40. A1: Correctly simplifies the final probability to \(\frac{1}{8}\) (or 0.125).

Rearrange the second equation to make \(y\) the subject: \(y = x + 2\). Substitute this expression for \(y\) into the first equation: \(2x^2 - x(x + 2) = 15\). Expand and simplify: \(2x^2 - x^2 - 2x = 15\), which simplifies to \(x^2 - 2x - 15 = 0\). Factorising the quadratic equation gives \((x - 5)(x + 3) = 0\). This yields two values for \(x\): \(x = 5\) or \(x = -3\). Substitute these values back into \(y = x + 2\) to find the corresponding values of \(y\). When \(x = 5\), \(y = 5 + 2 = 7\). When \(x = -3\), \(y = -3 + 2 = -1\). The solutions are \(x = 5, y = 7\) and \(x = -3, y = -1\).

M1: Rearranges the linear equation to make \(y\) or \(x\) the subject, e.g., \(y = x + 2\). M1: Substitutes this into the quadratic equation to obtain a single quadratic equation in one variable. M1: Solves the resulting quadratic equation \(x^2 - 2x - 15 = 0\) by factorisation to find \(x = 5\) and \(x = -3\). A1: Finds the corresponding \(y\) values \(y = 7\) and \(y = -1\). A1: Pairs the answers correctly as \((5, 7)\) and \((-3, -1)\).

The center of the circle is the origin \(O(0, 0)\). The radius of the circle connecting the center to the point of contact \(P(-3, 4)\) has gradient \(m = \frac{4 - 0}{-3 - 0} = -\frac{4}{3}\). Since the tangent line \(L\) is perpendicular to the radius at the point of contact, its gradient is the negative reciprocal: \(m = \frac{3}{4}\). Using the point-slope form with the point \(P(-3, 4)\), the equation of \(L\) is \(y - 4 = \frac{3}{4}(x - (-3))\), which is \(y - 4 = \frac{3}{4}(x + 3)\). Multiply both sides by 4 to eliminate the fraction: \(4(y - 4) = 3(x + 3)\), which expands to \(4y - 16 = 3x + 9\). Rearranging this into the required form gives \(4y - 3x = 25\).

M1: Finds the gradient of the radius to be \(-\frac{4}{3}\). M1: Uses the perpendicular gradient rule to find the gradient of the tangent as \(\frac{3}{4}\). M1: Sets up the equation of the line using the gradient \(\frac{3}{4}\) and the point \((-3, 4)\). M1: Multiplies through by 4 to clear the fraction. A1: Correctly expresses the final equation in the required form as \(4y - 3x = 25\) (or equivalent integer forms such as \(3x - 4y = -25\)).

Using the Sine Rule to find the length of side \(BC\): \(\frac{BC}{\sin(45^\circ)} = \frac{AB}{\sin(60^\circ)}\). Substituting the known values: \(\frac{BC}{\frac{\sqrt{2}}{2}} = \frac{2\sqrt{6}}{\frac{\sqrt{3}}{2}}\), which simplifies to \(BC = \frac{2\sqrt{6} \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} = \frac{\sqrt{12}}{\frac{\sqrt{3}}{2}} = \frac{2\sqrt{3}}{\frac{\sqrt{3}}{2}} = 4\text{ cm}\). Next, find the third angle \(\angle ABC = 180^\circ - (45^\circ + 60^\circ) = 75^\circ\). To find \(\sin(75^\circ)\), use the compound angle identity: \(\sin(75^\circ) = \sin(45^\circ + 30^\circ) = \sin(45^\circ)\cos(30^\circ) + \cos(45^\circ)\sin(30^\circ) = \left(\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2} \times \frac{1}{2}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}\). Finally, apply the area formula: \(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\angle ABC) = \frac{1}{2} \times 2\sqrt{6} \times 4 \times \frac{\sqrt{6} + \sqrt{2}}{4} = \sqrt{6}(\sqrt{6} + \sqrt{2}) = 6 + \sqrt{12} = 6 + 2\sqrt{3}\text{ cm}^2\).

M1: Uses the Sine Rule to find side \(BC\), e.g., \(\frac{BC}{\sin(45^\circ)} = \frac{2\sqrt{6}}{\sin(60^\circ)}\). M1: Obtains \(BC = 4\). M1: Calculates \(\angle ABC = 75^\circ\) and evaluates \(\sin(75^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}\). M1: Applies the triangle area formula \(\frac{1}{2} \times AB \times BC \times \sin(75^\circ)\). A1: Correctly simplifies the area to the exact form \(6 + 2\sqrt{3}\).

First, convert the distance to metres: \(1.5\text{ km} = 1500\text{ m}\). The total distance is the area of the trapezium under the velocity-time graph, with parallel sides of 50 seconds and 100 seconds, and height \(V\): \(\text{Area} = \frac{1}{2} \times (50 + 100) \times V\). This gives \(1500 = \frac{1}{2} \times 150 \times V\), so \(1500 = 75V\), which yields \(V = 20\text{ m/s}\). During the final 30 seconds, the velocity decreases uniformly from \(20\text{ m/s}\) to \(0\text{ m/s}\). The deceleration is \(\frac{\text{Change in velocity}}{\text{Time}} = \frac{20}{30} = \frac{2}{3}\text{ m/s}^2\).

M1: Converts 1.5 km to 1500 m. M1: Sets up the area equation for the distance, \(\frac{1}{2}(50 + 100)V = 1500\). M1: Correctly calculates \(V = 20\text{ m/s}\). M1: Calculates deceleration as \(\frac{V}{30}\). A1: Correctly simplifies the final deceleration to \(\frac{2}{3}\).

Multiply both sides of the equation by the common denominator \((x - 2)(x + 1)\) to clear the fractions: \(x(x + 1) - 3(x - 2) = 2(x - 2)(x + 1)\). Expand both sides: the left side becomes \(x^2 + x - 3x + 6 = x^2 - 2x + 6\), and the right side becomes \(2(x^2 - x - 2) = 2x^2 - 2x - 4\). Equating these gives \(x^2 - 2x + 6 = 2x^2 - 2x - 4\). Rearranging the equation by subtracting \(x^2 - 2x + 6\) from both sides gives \(0 = x^2 - 10\). This simplifies to \(x^2 = 10\). Taking the square root of both sides gives the exact solutions \(x = \pm\sqrt{10}\).

M1: Multiplies through by the common denominator \((x - 2)(x + 1)\). M1: Correctly expands the left and right sides to get \(x^2 - 2x + 6\) and \(2x^2 - 2x - 4\). M1: Re-arranges the quadratic terms to form \(x^2 - 10 = 0\). M1: Recognises the next step is to solve \(x^2 = 10\). A1: Finds both exact solutions \(x = \pm\sqrt{10}\).

Let the total number of counters in the bag be \(x + 5\).
Since two counters are selected without replacement, the probability that both counters are blue is given by:
\(\text{P}(\text{Blue, Blue}) = \frac{x}{x+5} \times \frac{x-1}{x+4}\)

We are given that this probability is \(\frac{3}{11}\):
\(\frac{x(x-1)}{(x+5)(x+4)} = \frac{3}{11}\)

Expand the brackets:
\(\frac{x^2 - x}{x^2 + 9x + 20} = \frac{3}{11}\)

Multiply both sides by \(11(x^2 + 9x + 20)\):
\(11(x^2 - x) = 3(x^2 + 9x + 20)\)
\(11x^2 - 11x = 3x^2 + 27x + 60\)

Rearrange into a quadratic equation set to zero:
\(8x^2 - 38x - 60 = 0\)

Divide the entire equation by 2:
\(4x^2 - 19x - 30 = 0\)

Factorise the quadratic expression:
\((4x + 5)(x - 6) = 0\)

Since \(x\) must be a positive integer, we reject the solution \(x = -\frac{5}{4}\) and accept:
\(x = 6\)

There are 6 blue counters and 5 red counters, making a total of 11 counters.

The probability of selecting at least one red counter is the complement of selecting zero red counters (i.e., selecting two blue counters):
\(\text{P}(\text{at least one red}) = 1 - \text{P}(\text{both blue}) = 1 - \frac{3}{11} = \frac{8}{11}\).

M1: Sets up the probability equation for both counters being blue: \(\frac{x}{x+5} \times \frac{x-1}{x+4} = \frac{3}{11}\).
M1: Correctly expands and rearranges the equation to form a quadratic, e.g., \(8x^2 - 38x - 60 = 0\) or \(4x^2 - 19x - 30 = 0\).
M1: Correctly factorises their quadratic to find the positive integer solution, obtaining \(x = 6\).
M1: Recognises that the probability of at least one red is \(1 - \text{P}(\text{both blue})\) or calculates \(\text{P}(\text{Red, Blue}) + \text{P}(\text{Blue, Red}) + \text{P}(\text{Red, Red})\).
A1 (accuracy): Obtains the correct final probability of \(\frac{8}{11}\) (or equivalent fraction).

Step 1: Find the coordinates of the intersection points \(A\) and \(B\) by setting the equations equal to each other:
\(x^2 - 4x - 5 = 2x + 2\)
\(x^2 - 6x - 7 = 0\)

Factorise the quadratic equation:
\((x - 7)(x + 1) = 0\)

This gives \(x = 7\) and \(x = -1\).

Find the corresponding \(y\)-coordinates using \(y = 2x + 2\):
- For \(x = -1\): \(y = 2(-1) + 2 = 0\), so \(A = (-1, 0)\).
- For \(x = 7\): \(y = 2(7) + 2 = 16\), so \(B = (7, 16)\).

Step 2: Find the coordinates of the minimum point \(V\) of the curve \(C\).
We can complete the square for \(y = x^2 - 4x - 5\):
\(y = (x - 2)^2 - 4 - 5\)
\(y = (x - 2)^2 - 9\)

So the minimum point (vertex) is \(V = (2, -9)\).

Step 3: Calculate the area of triangle \(ABV\) with vertices \(A(-1, 0)\), \(B(7, 16)\), and \(V(2, -9)\).
We can draw a bounding box (rectangle) around the triangle from \(x = -1\) to \(x = 7\) (width \(8\)) and from \(y = -9\) to \(y = 16\) (height \(25\)).

Area of bounding box = \(8 \times 25 = 200\).

Now, subtract the areas of the three right-angled triangles outside \(ABV\) within this bounding box:
1. Triangle 1 (bottom-left, between \(A(-1,0)\), \(V(2,-9)\), and \((-1,-9)\)):
Width = \(2 - (-1) = 3\), Height = \(0 - (-9) = 9\).
Area 1 = \(\frac{1}{2} \times 3 \times 9 = 13.5\).

2. Triangle 2 (bottom-right, between \(V(2,-9)\), \(B(7,16)\), and \((7,-9)\)):
Width = \(7 - 2 = 5\), Height = \(16 - (-9) = 25\).
Area 2 = \(\frac{1}{2} \times 5 \times 25 = 62.5\).

3. Triangle 3 (top-left, between \(A(-1,0)\), \(B(7,16)\), and \((-1,16)\)):
Width = \(7 - (-1) = 8\), Height = \(16 - 0 = 16\).
Area 3 = \(\frac{1}{2} \times 8 \times 16 = 64\).

Total area to subtract = \(13.5 + 62.5 + 64 = 140\).

Area of triangle \(ABV = 200 - 140 = 60\).

M1: Equates the line and curve equations to form a single quadratic: \(x^2 - 6x - 7 = 0\).
M1: Factorises and solves to find both \(x\)-coordinates: \(x = -1\) and \(x = 7\).
M1: Determines the full coordinates of \(A(-1, 0)\) and \(B(7, 16)\).
M1: Identifies the vertex/minimum point \(V(2, -9)\) by completing the square or differentiation.
M1: Standard method to find the area of a triangle given three coordinates (e.g., bounding box subtraction or Shoelace formula).
A1 (accuracy): Obtains the correct area of \(60\).

By Pythagoras' theorem, the sum of the squares of the two shorter sides equals the square of the hypotenuse:
\((x + 2)^2 + (x + 7)^2 = (x + 9)^2\)

Expand the algebraic expressions:
\((x^2 + 4x + 4) + (x^2 + 14x + 49) = x^2 + 18x + 81\)

Simplify and combine like terms on the left side:
\(2x^2 + 18x + 53 = x^2 + 18x + 81\)

Subtract \(x^2 + 18x + 53\) from both sides:
\(x^2 = 28\)

Since side lengths must be positive, \(x\) must be positive:
\(x = \sqrt{28} = 2\sqrt{7}\)

Now, substitute \(x = 2\sqrt{7}\) into the expressions for the three sides to find their lengths:
- Side 1: \(x + 2 = 2\sqrt{7} + 2\)
- Side 2: \(x + 7 = 2\sqrt{7} + 7\)
- Hypotenuse: \(x + 9 = 2\sqrt{7} + 9\)

Calculate the perimeter by summing the three sides:
\(\text{Perimeter} = (2\sqrt{7} + 2) + (2\sqrt{7} + 7) + (2\sqrt{7} + 9)\)
\(\text{Perimeter} = (2 + 7 + 9) + (2\sqrt{7} + 2\sqrt{7} + 2\sqrt{7})\)
\(\text{Perimeter} = 18 + 6\sqrt{7}\) cm.

M1: Applies Pythagoras' theorem: \((x+2)^2 + (x+7)^2 = (x+9)^2\).
M1: Correctly expands all quadratic terms to get \(x^2 + 4x + 4 + x^2 + 14x + 49 = x^2 + 18x + 81\).
M1: Simplifies the quadratic equation to \(x^2 = 28\).
M1: Obtains \(x = \sqrt{28}\) or \(2\sqrt{7}\).
M1: Sums the expressions for the side lengths to get \(3x + 18\).
A1 (accuracy): Plugs in the value of \(x\) to find the correct perimeter of \(18 + 6\sqrt{7}\).

Step 1: Use the area of a triangle formula to find the length of side \(BC\).
\(\text{Area} = \frac{1}{2} a c \sin B\)
\(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(150^\circ)\)

We know that:
\(\sin(150^\circ) = \sin(180^\circ - 150^\circ) = \sin(30^\circ) = \frac{1}{2}\)

Substitute the given values into the formula:
\(3\sqrt{3} = \frac{1}{2} \times 2\sqrt{3} \times BC \times \frac{1}{2}\)
\(3\sqrt{3} = \frac{\sqrt{3}}{2} \times BC\)

Divide both sides by \(\sqrt{3}\):
\(3 = \frac{1}{2} \times BC\)
\(BC = 6\) cm.

Step 2: Use the Cosine Rule to find the length of side \(AC\).
\(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(150^\circ)\)

We know that:
\(\cos(150^\circ) = -\cos(30^\circ) = -\frac{\sqrt{3}}{2}\)

Substitute the values into the Cosine Rule:
\(AC^2 = (2\sqrt{3})^2 + 6^2 - 2 \times 2\sqrt{3} \times 6 \times \left(-\frac{\sqrt{3}}{2}\right)\)
\(AC^2 = (4 \times 3) + 36 + 24\sqrt{3} \times \frac{\sqrt{3}}{2}\)
\(AC^2 = 12 + 36 + 12 \times 3\)
\(AC^2 = 48 + 36\)
\(AC^2 = 84\)

Take the square root of both sides:
\(AC = \sqrt{84} = \sqrt{4 \times 21} = 2\sqrt{21}\) cm.

M1: States or uses \(\text{Area} = \frac{1}{2} a c \sin B\).
M1: Substitutes the known values into the area formula, using \(\sin(150^\circ) = \frac{1}{2}\).
M1: Correctly calculates \(BC = 6\).
M1: Recalls and states the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(150^\circ)\).
M1: Substitutes the values including \(\cos(150^\circ) = -\frac{\sqrt{3}}{2}\).
M1: Evaluates to find \(AC^2 = 84\).
A1 (accuracy): Simplifies the surd to obtain the exact length \(2\sqrt{21}\).

The total number of marbles in the box is \(n + 4\).
The probability of selecting a blue marble on the first draw is \(\frac{4}{n+4}\).
If a blue marble is chosen, the probability of selecting another blue marble on the second draw is \(\frac{3}{n+3}\).

The combined probability of selecting two blue marbles is:
\(\frac{4}{n+4} \times \frac{3}{n+3} = \frac{12}{(n+4)(n+3)}\)

We are given that this probability is \(\frac{1}{11}\):
\(\frac{12}{(n+4)(n+3)} = \frac{1}{11}\)

Cross-multiplying gives:
\(12 \times 11 = (n+4)(n+3)\)
\(132 = n^2 + 7n + 12\)

Rearranging into a quadratic equation equal to zero:
\(n^2 + 7n - 120 = 0\)

Factorising the quadratic equation:
\((n + 15)(n - 8) = 0\)

Since the number of red marbles \(n\) must be positive, we choose \(n = 8\).

M1: For a correct probability equation in terms of \(n\), e.g., \(\frac{4}{n+4} \times \frac{3}{n+3} = \frac{1}{11}\)
M1: For expanding and forming a three-term quadratic equation, e.g., \(n^2 + 7n - 120 = 0\)
A1: For \(n = 8\) (rejecting \(n = -15\))

Set the two equations equal to each other to find the \(x\)-coordinates of the points of intersection:
\(2x^2 - 5x + 4 = 2x + 1\)

Subtract \(2x + 1\) from both sides to form a quadratic equation:
\(2x^2 - 7x + 3 = 0\)

Factorise the quadratic:
\((2x - 1)(x - 3) = 0\)

This gives:
\(x = 0.5\) or \(x = 3\)

Substitute these \(x\)-values back into the linear equation \(y = 2x + 1\) to find the corresponding \(y\)-values:
For \(x = 0.5\):
\(y = 2(0.5) + 1 = 2\)

For \(x = 3\):
\(y = 2(3) + 1 = 7\)

Thus, the points of intersection are \((0.5, 2)\) and \((3, 7)\).

M1: Set up \(2x^2 - 5x + 4 = 2x + 1\) and rearrange to form a quadratic equal to 0, e.g., \(2x^2 - 7x + 3 = 0\)
M1: Solve the quadratic to find both \(x\)-values: \(x = 0.5\) and \(x = 3\)
A1: Correct coordinates: \((0.5, 2)\) and \((3, 7)\)

To clear the fractions, multiply all terms by the common denominator \((x+2)(x-1)\):
\(5(x-1) + 3(x+2) = 2(x+2)(x-1)\)

Expand the brackets:
\(5x - 5 + 3x + 6 = 2(x^2 + x - 2)\)
\(8x + 1 = 2x^2 + 2x - 4\)

Rearrange into the standard form \(ax^2 + bx + c = 0\):
\(2x^2 - 6x - 5 = 0\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(-5)}}{2(2)}
\)x = \frac{6 \pm \sqrt{36 + 40}}{4}
\(x = \frac{6 \pm \sqrt{76}}{4}

Calculating the decimal values:
\)x = \frac{6 + 8.717798...}{4} \approx 3.68\)
\(x = \frac{6 - 8.717798...}{4} \approx -0.68\)

M1: For removing fractions correctly to obtain \(5(x-1) + 3(x+2) = 2(x+2)(x-1)\) or equivalent
M1: For simplifying to form a three-term quadratic equation, e.g., \(2x^2 - 6x - 5 = 0\)
A1: For both correct solutions: \(x = 3.68\) and \(x = -0.68\) (correct to 2 decimal places)

Let \(x\) be the number of students who like both Rock and Pop music.
Using the principle of inclusion-exclusion, the total number of students can be written as:
\(\text{Rock only} + \text{Both} + \text{Pop only} + \text{Neither} = 80\)
\((45 - x) + x + (38 - x) + 12 = 80\)

Simplifying the equation:
\(95 - x = 80\)
\(x = 15\)

So, 15 students like both Rock and Pop.

The number of students who like Rock music but not Pop music is:
\(45 - x = 45 - 15 = 30\)

The probability of selecting a student who likes Rock but not Pop is:
\(\frac{30}{80} = \frac{3}{8} = 0.375\)

M1: For an algebraic equation or Venn diagram setup to find the intersection, e.g., \((45 - x) + x + (38 - x) + 12 = 80\)
M1: For calculating the number of students who like Rock but not Pop as 30
A1: For the correct probability \(\frac{3}{8}\) or equivalent decimal \(0.375\) (or percentage \(37.5\%\))

First, find the gradient of the radius from the origin \((0,0)\) to the point \(P(3, 5)\):
\(\text{Gradient of radius} = \frac{5 - 0}{3 - 0} = \frac{5}{3}\)

Since the tangent is perpendicular to the radius, the gradient of the tangent is the negative reciprocal of \(\frac{5}{3}\):
\(\text{Gradient of tangent} = -\frac{3}{5}\)

Using the equation of a straight line with point \(P(3, 5)\) and gradient \(-\frac{3}{5}\):
\(y - 5 = -\frac{3}{5}(x - 3)\)

Multiply both sides by 5 to clear the fraction:
\(5(y - 5) = -3(x - 3)\)
\(5y - 25 = -3x + 9\)

Rearranging into the form \(ay + bx = c\):
\(5y + 3x = 34\)

M1: For finding the gradient of the radius to be \(\frac{5}{3}\) and stating or using the perpendicular gradient rule to get the tangent's gradient of \(-\frac{3}{5}\)
M1: For substituting their perpendicular gradient and the point \((3, 5)\) into a linear equation, e.g., \(y - 5 = -\frac{3}{5}(x - 3)\)
A1: For the final equation in the required integer form, e.g., \(5y + 3x = 34\) (or equivalent with integer coefficients)

Substitute the linear equation for \(y\) into the quadratic equation:
\(x^2 + (2x - 5)^2 = 25\)

Expand the brackets:
\(x^2 + 4x^2 - 20x + 25 = 25\)

Collect like terms:
\(5x^2 - 20x + 25 = 25\)

Subtract 25 from both sides to form a quadratic equation equal to zero:
\(5x^2 - 20x = 0\)

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

This gives two values for \(x\):
\(x = 0\) or \(x = 4\)

Substitute these back into \(y = 2x - 5\) to find the corresponding values of \(y\):
For \(x = 0\):
\(y = 2(0) - 5 = -5\)

For \(x = 4\):
\(y = 2(4) - 5 = 3\)

So the solutions are \(x = 0, y = -5\) and \(x = 4, y = 3\).

M1: For substituting \(y = 2x - 5\) into the circle equation to obtain \(x^2 + (2x - 5)^2 = 25\)
M1: For expanding and simplifying to a solvable quadratic equation, e.g., \(5x^2 - 20x = 0\)
A1: For both pairs of correct answers: \(x = 0, y = -5\) and \(x = 4, y = 3\)

The total distance travelled is equal to the area under the speed-time graph.
The graph has two sections:
1. A triangular acceleration phase from \(t = 0\) to \(t = 4\):
\(\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times v = 2v\)

2. A rectangular constant speed phase from \(t = 4\) to \(t = 12\):
\(\text{Area}_2 = \text{width} \times \text{height} = (12 - 4) \times v = 8v\)

The total area is:
\(\text{Total Area} = 2v + 8v = 10v\)

Since the total distance is 120 metres:
\(10v = 120\)
\(v = 12\)

M1: For an expression representing the area of either section, i.e., \(2v\) or \(8v\)
M1: For setting up the total area equation, e.g., \(2v + 8v = 120\) or \(\frac{1}{2}(8 + 12)v = 120\)
A1: For \(v = 12\)

We use the Cosine Rule to calculate the side length \(AC\) (let \(AC = b\), \(BC = a = 10\), and \(AB = c = 7\)):
\(b^2 = a^2 + c^2 - 2ac \cos(B)\)

Substitute the known values into the equation:
\(AC^2 = 10^2 + 7^2 - 2(10)(7)\cos(60^\circ)\)

Since \(\cos(60^\circ) = 0.5\):
\(AC^2 = 100 + 49 - 140(0.5)\)
\(AC^2 = 149 - 70\)
\(AC^2 = 79\)

Take the square root of both sides:
\(AC = \sqrt{79} \approx 8.88819... \text{ cm}\)

Rounding to 3 significant figures gives \(8.89\text{ cm}\).

M1: For substituting correctly into the Cosine Rule formula, e.g., \(10^2 + 7^2 - 2(10)(7)\cos(60^\circ)\)
M1: For evaluating to \(AC^2 = 79\) or \(AC = \sqrt{79}\)
A1: For \(8.89\) (accept answers in the range \(8.88\) to \(8.90\))

Let \(n\) be the number of red counters in the bag. The probability of selecting a red counter first is \(\frac{n}{8}\). Since the selection is without replacement, the probability of selecting a red counter second is \(\frac{n-1}{7}\). The probability that both counters are red is given by \(\frac{n}{8} \times \frac{n-1}{7} = \frac{3}{14}\). Multiplying both sides by 56 to clear the denominators gives \(n(n-1) = \frac{3}{14} \times 56\) which simplifies to \(n(n-1) = 12\). This gives the quadratic equation \(n^2 - n - 12 = 0\). Factoring the quadratic equation gives \((n - 4)(n + 3) = 0\). Since \(n\) must be a positive integer, we discard \(n = -3\). Thus, \(n = 4\).

M1: For setting up the probability equation: \(\frac{n}{8} \times \frac{n-1}{7} = \frac{3}{14}\) (or equivalent). M1: For simplifying to a quadratic equation, e.g., \(n^2 - n - 12 = 0\). A1: For 4 (reject -3).

The coordinates of the minimum point of a quadratic curve of the form \(y = (x - h)^2 + k\) are \((h, k)\). Given the minimum point is \((3, -2)\), we can write the equation of the curve in completed square form as \(y = (x - 3)^2 - 2\). Expanding this expression: \(y = (x^2 - 6x + 9) - 2\), which simplifies to \(y = x^2 - 6x + 7\). Comparing this with \(y = x^2 + bx + c\), we get \(b = -6\) and \(c = 7\).

M1: For writing the equation in the completed square form \((x-3)^2 - 2\) (or equivalent). M1: For expanding the brackets correctly to get \(x^2 - 6x + 9 - 2\) (allow one sign error). A1: For both \(b = -6\) and \(c = 7\) correct.

First, use the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a b \sin C\). This gives \(28 = \frac{1}{2} \times 7 \times 10 \times \sin(\angle ABC)\), which simplifies to \(28 = 35 \sin(\angle ABC)\), so \(\sin(\angle ABC) = \frac{28}{35} = 0.8\). Since angle \(ABC\) is obtuse (\(90^\circ < \angle ABC < 180^\circ\)), we find \(\angle ABC = 180^\circ - \sin^{-1}(0.8) \approx 126.87^\circ\) (or \(\cos(\angle ABC) = -0.6\)). Now, use the cosine rule to find the length of \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle ABC)\). Substituting the values gives \(AC^2 = 7^2 + 10^2 - 2 \times 7 \times 10 \times \cos(126.87^\circ) = 49 + 100 - 140 \times (-0.6) = 149 + 84 = 233\). Thus, \(AC = \sqrt{233} \approx 15.264\text{ cm}\). To 3 significant figures, \(AC = 15.3\text{ cm}\).

M1: For using the area formula to find \(\sin(\angle ABC) = 0.8\) (or equivalent equation). M1: For finding \(\angle ABC \approx 126.9^\circ\) (or \(\cos(\angle ABC) = -0.6\)) and substituting into the cosine rule formula. A1: For 15.3 (accept answers in the range 15.26 to 15.3).

The total number of counters in the bag is \(8 + n\).

The probability of choosing a red counter first is \(\frac{8}{8+n}\).

Since the first counter is not replaced, there are now 7 red counters remaining out of a total of \(7 + n\) counters.

The probability of choosing a red counter second is \(\frac{7}{7+n}\).

The probability that both are red is:
\(\frac{8}{8+n} \times \frac{7}{7+n} = \frac{14}{39}\)

\(\frac{56}{(8+n)(7+n)} = \frac{14}{39}\)

Divide both sides of the equation by 14:
\(\frac{4}{(8+n)(7+n)} = \frac{1}{39}\)

Cross-multiplying gives:
\((8+n)(7+n) = 156\)

\(56 + 15n + n^2 = 156\)

\(n^2 + 15n - 100 = 0\)

Factorising the quadratic equation:
\((n + 20)(n - 5) = 0\)

Since \(n\) must be a positive integer, \(n = 5\).

M1: For writing an expression for the probability of selecting two red counters without replacement: \(\frac{8}{8+n} \times \frac{7}{7+n}\) (or equivalent).
M1: For setting up the equation \(\frac{56}{(8+n)(7+n)} = \frac{14}{39}\) and attempting to expand and rearrange.
M1: For obtaining a correct quadratic equation, e.g., \(n^2 + 15n - 100 = 0\).
M1 (dep): For solving the quadratic equation to find \(n = 5\) and \(n = -20\).
A0.5: For concluding that \(n = 5\) (rejecting \(n = -20\)).

The centre of the circle is at the origin \(O(0, 0)\).

The gradient of the radius \(OP\) is:
\(m_{\text{radius}} = \frac{5 - 0}{-3 - 0} = -\frac{5}{3}\).

Since the tangent is perpendicular to the radius, the gradient of the tangent is:
\(m_{\text{tangent}} = -\frac{1}{-\frac{5}{3}} = \frac{3}{5} = 0.6\).

Using the equation of a straight line \(y - y_1 = m(x - x_1)\) with point \(P(-3, 5)\):
\(y - 5 = 0.6(x - (-3))\)
\(y - 5 = 0.6(x + 3)\)
\(y - 5 = 0.6x + 1.8\)
\(y = 0.6x + 6.8\)

M1: For calculating the gradient of the radius \(OP\): \(\frac{5}{-3} = -\frac{5}{3}\).
M1: For using the perpendicular gradient rule \(m_1 m_2 = -1\) to find the gradient of the tangent: \(m = \frac{3}{5}\) (or 0.6).
M1: For substituting the point \((-3, 5)\) and their perpendicular gradient into a line equation formula, e.g., \(y - 5 = m(x + 3)\).
A1.5: For the correct final equation \(y = 0.6x + 6.8\) or \(y = \frac{3}{5}x + \frac{34}{5}\) (or any equivalent form matching \(y = mx + c\)).

Multiply the entire equation by the common denominator \(x(x+3)\):
\(18(x+3) - 18x = 1 \cdot x(x+3)\)

Expand the brackets:
\(18x + 54 - 18x = x^2 + 3x\)
\(54 = x^2 + 3x\)

Rearrange into standard quadratic form:
\(x^2 + 3x - 54 = 0\)

Factorise the quadratic expression:
\((x + 9)(x - 6) = 0\)

This gives the solutions:
\(x = 6\) or \(x = -9\).

M1: For a correct attempt to clear denominators by multiplying by \(x(x+3)\): \(18(x+3) - 18x = x(x+3)\).
M1: For expanding brackets correctly: \(18x + 54 - 18x = x^2 + 3x\).
M1: For rearranging to a standard quadratic form \(x^2 + 3x - 54 = 0\).
M1: For attempting to solve their quadratic equation by factorising, completing the square, or using the formula.
A0.5: For both correct solutions: \(x = 6\) and \(x = -9\).

Let's split the journey into three stages:
- **Stage 1 (Acceleration):**
Time \(t_1 = 10\) seconds.
Distance \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times v = 5v\).

- **Stage 2 (Constant Speed):**
Time \(t_2 = 15\) seconds.
Distance \(d_2 = \text{base} \times \text{height} = 15 \times v = 15v\).

- **Stage 3 (Deceleration):**
Deceleration \(a = 0.8\text{ m/s}^2\).
Time taken to stop from speed \(v\) is \(t_3 = \frac{v}{0.8} = 1.25v\) seconds.
Distance \(d_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1.25v \times v = 0.625v^2\).

The total distance is:
\(5v + 15v + 0.625v^2 = 200\)
\(0.625v^2 + 20v - 200 = 0\)

Multiply by 8 to clear decimals:
\(5v^2 + 160v - 1600 = 0\)

Divide by 5:
\(v^2 + 32v - 320 = 0\)

Factorising:
\((v - 8)(v + 40) = 0\)

Since speed \(v\) must be positive, \(v = 8\).

M1: For expressing the distance of Stage 1 as \(5v\) and Stage 2 as \(15v\).
M1: For setting up the time of Stage 3 as \(\frac{v}{0.8}\) and finding the distance as \(\frac{1}{2} \times v \times \frac{v}{0.8} = 0.625v^2\).
M1: For setting up the total distance equation: \(20v + 0.625v^2 = 200\).
M1: For solving the quadratic equation to find \(v = 8\) and \(v = -40\).
A0.5: For identifying \(v = 8\) as the final positive speed.

First, find the length of side \(PR\) using the Cosine Rule:
\(PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(PQR)\)
\(PR^2 = 7.4^2 + 5.8^2 - 2 \times 7.4 \times 5.8 \times \cos(105^\circ)\)
\(PR^2 = 54.76 + 33.64 - 85.84 \times (-0.258819)\)
\(PR^2 \approx 88.40 + 22.217 = 110.617\)
\(PR \approx \sqrt{110.617} \approx 10.517\text{ cm}\).

Next, find angle \(QPR\) (let's call it \(\theta\)) using the Sine Rule:
\(\frac{\sin(\theta)}{QR} = \frac{\sin(PQR)}{PR}\)
\(\sin(\theta) = \frac{5.8 \times \sin(105^\circ)}{10.517}\)
\(\sin(\theta) \approx \frac{5.8 \times 0.965926}{10.517} \approx \frac{5.60237}{10.517} \approx 0.5327\)
\(\theta \approx \arcsin(0.5327) \approx 32.189^\circ\).

To 1 decimal place, the angle is \(32.2^\circ\).

M1: For substituting the given values correctly into the Cosine Rule: \(PR^2 = 7.4^2 + 5.8^2 - 2 \times 7.4 \times 5.8 \times \cos(105^\circ)\).
M1: For calculating \(PR^2 \approx 110.617\) or \(PR \approx 10.5\text{ cm}\).
M1: For substituting into the Sine Rule to find \(\sin(\theta)\): \(\frac{\sin(\theta)}{5.8} = \frac{\sin(105^\circ)}{10.517}\).
A1.5: For the final angle 32.2 (accept answers in range 32.1 to 32.3).

We are selecting from the subset of students who study Chemistry.

The total number of Chemistry students is 65.

We need to find how many of these Chemistry students study Biology but not Physics. This is the region \(C \cap B \cap P'\).

We know:
- The total number of students who study both Chemistry and Biology is 25.
- The number of students who study all three subjects (Chemistry, Biology, and Physics) is 10.

Therefore, the number of students who study Chemistry and Biology but not Physics is:
\(25 - 10 = 15\) students.

Since we are selecting two different students without replacement from the 65 Chemistry students:
- Probability that the first student studies Chemistry and Biology but not Physics is \(\frac{15}{65} = \frac{3}{13}\).
- Probability that the second student also does is \(\frac{14}{64} = \frac{7}{32}\).

The joint probability is:
\(\frac{3}{13} \times \frac{7}{32} = \frac{21}{416}\).

M1: For identifying or calculating the number of Chemistry students who study Biology but not Physics: \(25 - 10 = 15\) students.
M1: For setting up the first probability as \(\frac{15}{65}\) (or \(\frac{3}{13}\)).
M1: For multiplying their first probability by a correct second probability representing selection without replacement, e.g., \(\frac{14}{64}\) (or \(\frac{7}{32}\)).
A1.5: For simplifying to the correct final fraction \(\frac{21}{416}\).

The total number of counters in the bag is \(x + (x+2) + 6 = 2x+8\). The probability of drawing two counters of the same colour is the sum of drawing two red, two blue, or two green. This is given by \(P(\text{same}) = \frac{x(x-1)}{(2x+8)(2x+7)} + \frac{(x+2)(x+1)}{(2x+8)(2x+7)} + \frac{6 \times 5}{(2x+8)(2x+7)} = \frac{29}{95}\). Expanding and simplifying the numerators gives \(x^2 - x + x^2 + 3x + 2 + 30 = 2x^2 + 2x + 32\). The denominator is \((2x+8)(2x+7) = 4x^2 + 30x + 56\). Setting up the equation: \(\frac{2x^2 + 2x + 32}{4x^2 + 30x + 56} = \frac{29}{95}\). Cross-multiplying: \(95(2x^2 + 2x + 32) = 29(4x^2 + 30x + 56)\), which expands to \(190x^2 + 190x + 3040 = 116x^2 + 870x + 1624\). Rearranging gives \(74x^2 - 680x + 1416 = 0\). Dividing by 2, we get \(37x^2 - 340x + 708 = 0\). Solving the quadratic using the formula: \(x = \frac{340 \pm \sqrt{(-340)^2 - 4(37)(708)}}{2(37)} = \frac{340 \pm \sqrt{115600 - 104784}}{74} = \frac{340 \pm 104}{74}\). This gives \(x = 6\) or \(x \approx 3.19\). Since \(x\) must be an integer, \(x = 6\). With \(x = 6\), the bag contains 6 red, 8 blue, and 6 green counters, making a total of 20 counters. The probability of choosing at least one green counter is \(1 - P(\text{no green}) = 1 - \frac{14}{20} \times \frac{13}{19} = 1 - \frac{182}{380} = \frac{198}{380} = \frac{99}{190}\).

M1: Writes expressions for the probability of selecting two counters of each individual colour. M1: Forms the equation for the total probability of same colour. M1: Correctly expands and simplifies the numerator and denominator. A1: Correctly cross-multiplies to obtain a quadratic equation. A1: Shows the reduction to the given equation \(37x^2 - 340x + 708 = 0\). M1: Solves the quadratic equation to find possible values of \(x\). A1: Selects the correct integer value \(x=6\). M1: Calculates the total number of non-green counters (14) and the total counters (20). M1: Forms the product for the probability of selecting no green counters. A1: Subtracts from 1 to find the probability of at least one green counter. A1: Correctly simplifies the final probability to \(\frac{99}{190}\) or equivalent decimal (\(\approx 0.521\)).

The time taken by Anita is \(\frac{10}{v}\) hours. The time taken by Beth is \(\frac{4}{v-2} + \frac{6}{v+5}\) hours. Since Beth finishes 6 minutes (which is \(\frac{6}{60} = \frac{1}{10}\) hours) before Anita, we have: \(\frac{10}{v} - (\frac{4}{v-2} + \frac{6}{v+5}) = \frac{1}{10}\). Multiplying the entire equation by the common denominator \(10v(v-2)(v+5)\) yields: \(100(v-2)(v+5) - 10v[4(v+5) + 6(v-2)] = v(v-2)(v+5)\). Expanding these terms: \(100(v^2 + 3v - 10) - 10v[4v + 20 + 6v - 12] = v(v^2 + 3v - 10)\), which simplifies to \(100v^2 + 300v - 1000 - 10v[10v + 8] = v^3 + 3v^2 - 10v\). This further simplifies to \(100v^2 + 300v - 1000 - 100v^2 - 80v = v^3 + 3v^2 - 10v\). Grouping the terms on one side: \(220v - 1000 = v^3 + 3v^2 - 10v \implies v^3 + 3v^2 - 230v + 1000 = 0\). Testing \(v = 10\) in the cubic equation: \(10^3 + 3(10^2) - 230(10) + 1000 = 1000 + 300 - 2300 + 1000 = 0\). Thus, \(v = 10\) is a solution. Dividing \(v^3 + 3v^2 - 230v + 1000\) by \(v-10\) yields the quadratic factor \(v^2 + 13v - 100 = 0\). Solving \(v^2 + 13v - 100 = 0\) using the quadratic formula gives \(v \approx 5.43\) or a negative speed. Since we are given \(v > 6\), the only valid solution for Anita's speed is \(v = 10\) km/h.

M1: Expresses Anita's time as \(\frac{10}{v}\). M1: Expresses Beth's time as \(\frac{4}{v-2} + \frac{6}{v+5}\). M1: Converts 6 minutes to \(\frac{1}{10}\) hours. M1: Sets up the difference equation \(\frac{10}{v} - (\frac{4}{v-2} + \frac{6}{v+5}) = \frac{1}{10}\). A1: Multiplies by the common denominator to clear algebraic fractions. A1: Correctly expands LHS terms. A1: Correctly expands RHS terms. A1: Shows simplification to the given cubic equation \(v^3 + 3v^2 - 230v + 1000 = 0\). M1: Identifies \(v=10\) as a root of the equation. M1: Divides by \((v-10)\) to find the quadratic factor \(v^2 + 13v - 100 = 0\). A1: Evaluates other roots to show they are either negative or less than 6. A1: Concludes that the unique valid speed is \(10\).

First, find the area of triangle \(ABD\): \(\text{Area}_{ABD} = \frac{1}{2} \times AB \times AD \times \sin(BAD) = \frac{1}{2} \times 120 \times 85 \times \sin(68^\circ) \approx 4728.64\text{ m}^2\). Next, find the length of the diagonal \(BD\) using the Cosine Rule on triangle \(ABD\): \(BD^2 = AB^2 + AD^2 - 2 \times AB \times AD \times \cos(BAD) = 120^2 + 85^2 - 2 \times 120 \times 85 \times \cos(68^\circ) = 14400 + 7225 - 20400 \times 0.374607 \approx 13983.03\). Therefore, \(BD = \sqrt{13983.03} \approx 118.25\text{ m}\). In triangle \(BCD\), the third angle is \(\angle BDC = 180^\circ - (42^\circ + 75^\circ) = 63^\circ\). Using the Sine Rule to find the length of \(CD\): \(\frac{CD}{\sin(42^\circ)} = \frac{BD}{\sin(75^\circ)} \implies CD = \frac{118.25 \times \sin(42^\circ)}{\sin(75^\circ)} \approx \frac{118.25 \times 0.669131}{0.965926} \approx 81.92\text{ m}\). Now, calculate the area of triangle \(BCD\): \(\text{Area}_{BCD} = \frac{1}{2} \times BD \times CD \times \sin(\angle BDC) = \frac{1}{2} \times 118.25 \times 81.92 \times \sin(63^\circ) \approx 4315.02\text{ m}^2\). The total area of the quadrilateral \(ABCD\) is \(\text{Area}_{ABD} + \text{Area}_{BCD} \approx 4728.64 + 4315.02 = 9043.66\text{ m}^2\). Rounding to 3 significant figures gives \(9040\text{ m}^2\).

M1: Recalls and applies the area formula \(\frac{1}{2}ab\sin C\) for triangle \(ABD\). A1: Calculates area of triangle \(ABD\) as \(\approx 4728.64\). M1: Recalls and applies the Cosine Rule to find \(BD^2\). A1: Evaluates \(BD^2 \approx 13983\) or \(BD \approx 118.25\). M1: Finds the third angle in triangle \(BCD\) as \(63^\circ\). M1: Applies the Sine Rule in triangle \(BCD\). A1: Evaluates \(CD \approx 81.9\). M1: Calculates the area of triangle \(BCD\) using \(\frac{1}{2} \times BD \times CD \times \sin(63^\circ)\) or equivalent. A1: Evaluates area of triangle \(BCD\) as \(\approx 4315\). M1: Adds the two areas together. A1: Correctly rounds the total area to 3 significant figures, obtaining \(9040\).