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Volume of a hemisphere: \(V = \frac{2}{3} \pi r^3\). Substituting \(r = 6\) gives \(V = \frac{2}{3} \pi (216) = 144\pi\text{ cm}^3\). Volume of a cone: \(V = \frac{1}{3} \pi R^2 h\). Since the volume remains the same: \(144\pi = \frac{1}{3} \pi R^2 (8)\), which simplifies to \(144 = \frac{8}{3} R^2\). Solving for \(R^2\) gives \(R^2 = 54\). Thus, \(R = \sqrt{54} \approx 7.35\text{ cm}\) (correct to 3 significant figures).

M1 for \(\frac{2}{3} \times \pi \times 6^3\) or \(144\pi\) (or \(452.389...\))
M1 for equating their volume to \(\frac{1}{3} \times \pi \times R^2 \times 8\) and solving for \(R^2\)
A1 for \(7.35\) (accept \(\sqrt{54}\) or \(3\sqrt{6}\))

Using set notation, the total number of students can be expressed as: \(n(B \text{ only}) + n(C \text{ only}) + n(B \cap C) + n(\text{neither}) = 30\). This is written in terms of \(x\) as: \((18 - 2x) + (15 - 2x) + 2x + x = 30\). Simplifying this linear equation gives: \(33 - x = 30\). Solving for \(x\) yields \(x = 3\).

M1 for expressing the sum of all regions in terms of \(x\): \((18-2x) + 2x + (15-2x) + x = 30\) or equivalent
M1 for simplifying the equation to \(33 - x = 30\) or equivalent algebraic step
A1 for \(3\)

First, factorise the quadratic numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Next, factorise the quadratic denominator using the difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). The algebraic fraction becomes: \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\). Cancelling the common factor \((2x + 1)\) from both numerator and denominator yields the simplified expression: \(\frac{x - 3}{2x - 1}\).

M1 for factorising the numerator to \((2x + 1)(x - 3)\)
M1 for factorising the denominator to \((2x - 1)(2x + 1)\)
A1 for \(\frac{x - 3}{2x - 1}\)

To eliminate \(y\), multiply the first equation by 3: \(9x + 6y = 12\). Multiply the second equation by 2: \(8x - 6y = 22\). Add the two equations together: \(17x = 34\), which gives \(x = 2\). Substitute \(x = 2\) into the first equation: \(3(2) + 2y = 4 \implies 6 + 2y = 4 \implies 2y = -2 \implies y = -1\).

M1 for a correct method to eliminate one variable (e.g., aligning coefficients of \(y\))
M1 for substituting their found value of \(x\) (or \(y\)) into one of the original equations
A1 for \(x = 2, y = -1\)

Since the measurements are correct to 1 decimal place, the range of error is \(\pm 0.05\text{ cm}\). The upper bound of the base is \(b_{\text{upper}} = 8.4 + 0.05 = 8.45\text{ cm}\). The upper bound of the height is \(h_{\text{upper}} = 6.2 + 0.05 = 6.25\text{ cm}\). The upper bound of the area is: \(A_{\text{upper}} = \frac{1}{2} \times b_{\text{upper}} \times h_{\text{upper}} = \frac{1}{2} \times 8.45 \times 6.25 = 26.40625\text{ cm}^2\).

M1 for identifying upper bound of \(b\) as \(8.45\) or upper bound of \(h\) as \(6.25\)
M1 for \(\frac{1}{2} \times 8.45 \times 6.25\)
A1 for \(26.40625\) (or \(26.4\))

The gradient of the given line is \(m_1 = 3\). The gradient of the perpendicular line is the negative reciprocal: \(m_2 = -\frac{1}{3}\). Using the equation of a straight line with gradient \(-\frac{1}{3}\) passing through \((6, 2)\): \(y - 2 = -\frac{1}{3}(x - 6)\). Simplifying this gives: \(y - 2 = -\frac{1}{3}x + 2 \implies y = -\frac{1}{3}x + 4\).

M1 for gradient of perpendicular line as \(-\frac{1}{3}\)
M1 for substituting \((6, 2)\) and their gradient into \(y = mx + c\) (or equivalent)
A1 for \(y = -\frac{1}{3}x + 4\)

The terms are \(3, 11, 23, 39\). The first differences are \(8, 12, 16\), and the second differences are constant at \(4\). This indicates a quadratic sequence of the form \(an^2 + bn + c\), where \(a = \frac{4}{2} = 2\). Subtracting \(2n^2\) from each term yields: for \(n=1\), \(3-2=1\); for \(n=2\), \(11-8=3\); for \(n=3\), \(23-18=5\); for \(n=4\), \(39-32=7\). The remaining linear sequence is \(1, 3, 5, 7, \dots\), which has the \(n\)-th term \(2n - 1\). Thus, the overall \(n\)-th term is \(2n^2 + 2n - 1\).

M1 for finding second difference is \(4\) or stating leading term is \(2n^2\)
M1 for subtracting \(2n^2\) from terms to find the linear sequence \(1, 3, 5, 7\) or establishing correct linear coefficients
A1 for \(2n^2 + 2n - 1\)

Join \(OA\) and \(OC\). Since \(TA\) and \(TC\) are tangents, the angles \(\angle OAT\) and \(\angle OCT\) are both \(90^\circ\). In quadrilateral \(OATC\), the sum of angles is \(360^\circ\), which gives: \(\angle AOC = 360^\circ - 90^\circ - 90^\circ - 48^\circ = 132^\circ\). The angle subtended by an arc at the centre is twice the angle subtended by it at any point on the major arc circumference. Therefore: \(\angle ADC = \frac{1}{2} \times \angle AOC = \frac{1}{2} \times 132^\circ = 66^\circ\).

M1 for recognizing tangent-radius angles as \(90^\circ\) (e.g. \(90 + 90\) shown or implied)
M1 for calculating obtuse angle \(AOC = 132^\circ\)
A1 for \(66\)

1. Factorize the numerator:
\(2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\)

2. Factorize the denominator as a difference of two squares:
\(4x^2 - 1 = (2x - 1)(2x + 1)\)

3. Divide the numerator by the denominator, cancelling the common factor \((2x + 1)\):
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\)

M1 for factorizing the numerator to \((2x + 1)(x - 3)\) (or equivalent)
M1 for factorizing the denominator to \((2x - 1)(2x + 1)\)
A1 for the final simplified answer \(\frac{x - 3}{2x - 1}\)

1. Find the upper bound of the length: Since the length is correct to the nearest metre, the boundary value is \(\pm 0.5\text{ m}\).
\(\text{Upper Bound of Length} = 85 + 0.5 = 85.5\text{ m}\).

2. Find the upper bound of the width: Since the width is correct to the nearest \(5\text{ m}\), the boundary value is \(\pm 2.5\text{ m}\).
\(\text{Upper Bound of Width} = 40 + 2.5 = 42.5\text{ m}\).

3. Calculate the upper bound of the area:
\(\text{Upper Bound of Area} = 85.5 \times 42.5 = 3612.5\text{ m}^2\).

B1 for \(85.5\) seen
B1 for \(42.5\) seen
B1 for \(3612.5\) (if no working shown, full marks are awarded for the correct final answer)

Let \(P\) be the original price.

Since the price is reduced by \(15\%\), the sale price is \(100\% - 15\% = 85\%\) of the original price.

\(0.85 \times P = 646\)

\(P = \frac{646}{0.85} = 760\)

So, the original price was \(\$760\).

M1 for equating \(85\%\) to \(646\) (e.g., \(0.85x = 646\))
M1 for \(646 \div 0.85\) or equivalent calculation
A1 for \(760\)

1. Find angle \(AXB\): Since points \(A\), \(X\), and \(C\) lie on a straight line (as \(AC\) is a diameter),
\(\text{Angle } AXB = 180^\circ - 78^\circ = 102^\circ\).

2. In \(\triangle ABX\), find angle \(ABX\):
\(\text{Angle } ABX = 180^\circ - (102^\circ + 32^\circ) = 180^\circ - 134^\circ = 46^\circ\).
Therefore, angle \(ABD = 46^\circ\).

3. Use circle theorems: Angles subtended by the same arc \(AD\) at the circumference are equal.
Therefore, \(\text{Angle } ACD = \text{Angle } ABD = 46^\circ\).

M1 for finding angle \(AXB = 102^\circ\) (or angle \(XBC = 44^\circ\))
M1 for calculating angle \(ABD\) (or angle \(ABX\)) \(= 46^\circ\) (or angle \(DBC = 44^\circ\))
A1 for \(46\)

1. Find the gradient of the given line by rearranging it into \(y = mx + c\) form:
\(3y = 2x + 9 \implies y = \frac{2}{3}x + 3\).
So, the gradient of the given line is \(m_1 = \frac{2}{3}\).

2. Find the gradient of the perpendicular line \(m_2\):
\(m_2 = -\frac{1}{m_1} = -\frac{3}{2} = -1.5\).

3. Use the point \((4, -1)\) to determine the \(y\)-intercept \(c\):
\(y = mx + c\)
\(-1 = -1.5(4) + c\)
\(-1 = -6 + c \implies c = 5\).

The equation of the perpendicular line is \(y = -1.5x + 5\) (or \(y = -\frac{3}{2}x + 5\)).

M1 for rearranging the line equation to find its gradient is \(\frac{2}{3}\) (or identifying the perpendicular gradient as \(-\frac{3}{2}\) or \(-1.5\))
M1 for substituting \((4, -1)\) and their perpendicular gradient into \(y = mx + c\) to find \(c\)
A1 for \(y = -1.5x + 5\) or \(y = -\frac{3}{2}x + 5\)

1. Write the expression for the magnitude of the vector:
\(|\mathbf{p}| = \sqrt{(2k)^2 + (k - 1)^2}\)

2. Equate this to the given magnitude and square both sides:
\((2k)^2 + (k - 1)^2 = 40\)
\(4k^2 + k^2 - 2k + 1 = 40\)
\(5k^2 - 2k - 39 = 0\)

3. Solve the quadratic equation by factoring:
\((5k + 13)(k - 3) = 0\)
This gives \(k = -\frac{13}{5}\) or \(k = 3\).

Since \(k > 0\), the value of \(k\) is \(3\).

M1 for writing \((2k)^2 + (k-1)^2 = 40\) (or equivalent)
M1 for expanding and simplifying to a quadratic equation, e.g., \(5k^2 - 2k - 39 = 0\)
A1 for \(3\) (rejecting the negative root)

Let \(x\) be the number of students who study both Chemistry and Physics, so \(n(C \cap P) = x\).

The number of students who study at least one of the two subjects is:
\(n(C \cup P) = 30 - 5 = 25\).

Using the formula for the union of two sets:
\(n(C \cup P) = n(C) + n(P) - n(C \cap P)\)
\(25 = 18 + 15 - x\)
\(25 = 33 - x\)
\(x = 8\).

So, 8 students study both subjects.

M1 for \(30 - 5 = 25\) (number of students in the union) or \(18 + 15 + 5 - 30\)
M1 for setting up the equation \(18 + 15 - x = 25\) (or equivalent Venn diagram representation)
A1 for \(8\)

1. Find the first differences between successive terms:
\(11 - 4 = 7\)
\(22 - 11 = 11\)
\(37 - 22 = 15\)
First differences: \(7, 11, 15\)

2. Find the second differences:
\(11 - 7 = 4\)
\(15 - 11 = 4\)
Since the second differences are constant, the sequence is quadratic with a general term of \(an^2 + bn + c\).
The coefficient \(a\) is half of the second difference:
\(a = \frac{4}{2} = 2\).

3. Subtract \(2n^2\) from each term of the sequence to find the linear part:
For \(n = 1\): \(4 - 2(1)^2 = 2\)
For \(n = 2\): \(11 - 2(2)^2 = 3\)
For \(n = 3\): \(22 - 2(3)^2 = 4\)
For \(n = 4\): \(37 - 2(4)^2 = 5\)

4. The resulting sequence \(2, 3, 4, 5, \dots\) is linear, with a first term of 2 and a common difference of 1. Its \(n\)-th term is \(n + 1\).

Combining the quadratic and linear parts, the \(n\)-th term of the original sequence is:
\(2n^2 + n + 1\).

M1 for finding the second difference is \(4\) or determining the \(n^2\) coefficient is \(2\)
M1 for attempting to find the remaining linear sequence \(n + 1\) (or solving a system of simultaneous equations for \(b\) and \(c\))
A1 for the correct expression \(2n^2 + n + 1\) (or equivalent)

1. Find the upper bound of the length:
Since the length is correct to the nearest \(5\text{ m}\), the boundary interval is \(5 / 2 = 2.5\text{ m}\).
\(\text{Upper Bound of Length} = 80 + 2.5 = 82.5\text{ m}\)

2. Find the upper bound of the width:
Since the width is correct to the nearest \(1\text{ m}\), the boundary interval is \(1 / 2 = 0.5\text{ m}\).
\(\text{Upper Bound of Width} = 45 + 0.5 = 45.5\text{ m}\)

3. Calculate the upper bound for the perimeter:
\(\text{Perimeter} = 2 \times (\text{Length} + \text{Width})\)
\(\text{Upper Bound of Perimeter} = 2 \times (82.5 + 45.5) = 2 \times 128 = 256\text{ m}\)

M1 for upper bound of length \(82.5\) or upper bound of width \(45.5\) seen.
M1 for \(2 \times (\text{their } 82.5 + \text{their } 45.5)\).
A1 for \(256\).

1. Factorise the numerator, \(2x^2 - 5x - 3\):
We look for numbers that multiply to \(2 \times (-3) = -6\) and add to \(-5\). These are \(-6\) and \(1\).
\(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\)

2. Factorise the denominator, \(4x^2 - 1\), which is a difference of two squares:
\(4x^2 - 1 = (2x)^2 - 1^2 = (2x - 1)(2x + 1)\)

3. Simplify the fraction by canceling the common factor \((2x + 1)\):
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\)

M1 for numerator factorised correctly: \((2x + 1)(x - 3)\).
M1 for denominator factorised correctly: \((2x - 1)(2x + 1)\).
A1 for \(\frac{x - 3}{2x - 1}\) or equivalent correct simplified form.

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 8\), and \(c = -5\):

\(x = \frac{-8 \pm \sqrt{8^2 - 4(3)(-5)}}{2(3)}\)
\(x = \frac{-8 \pm \sqrt{64 + 60}}{6}\)
\(x = \frac{-8 \pm \sqrt{124}}{6}\)

Since \(\sqrt{124} \approx 11.1355\):

\(x_1 = \frac{-8 + 11.1355}{6} \approx 0.5226 \rightarrow 0.52\)
\(x_2 = \frac{-8 - 11.1355}{6} \approx -3.1893 \rightarrow -3.19\)

The solutions are \(0.52\) and \(-3.19\).

M1 for substitution of \(a, b, c\) correctly into the quadratic formula (allow one sign error in substitution).
M1 for \(\frac{-8 \pm \sqrt{124}}{6}\) or equivalent.
A1 for \(0.52\) and \(-3.19\) (both correct to 2 d.p.).

1. Express \(2\vec{a} - 3\vec{b}\) as a single column vector:
\(2\vec{a} = 2 \begin{pmatrix} -3 \\ 5 \end{pmatrix} = \begin{pmatrix} -6 \\ 10 \end{pmatrix}\)
\(3\vec{b} = 3 \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ -3
\end{pmatrix}\)

\(2\vec{a} - 3\vec{b} = \begin{pmatrix} -6 \\ 10 \end{pmatrix} - \begin{pmatrix} 6 \\ -3 \end{pmatrix} = \begin{pmatrix} -6 - 6 \\ 10 - (-3) \end{pmatrix} = \begin{pmatrix} -12 \\ 13 \end{pmatrix}\)

2. Find the magnitude of this vector:
\(|2\vec{a} - 3\vec{b}| = \sqrt{(-12)^2 + 13^2} = \sqrt{144 + 169} = \sqrt{313}\)
\(\sqrt{313} \approx 17.6918...\)

Rounding to 3 significant figures gives \(17.7\).

M1 for \(2\vec{a} - 3\vec{b} = \begin{pmatrix} -12 \\ 13 \end{pmatrix}\) or for finding either correct component.
M1 for \(\sqrt{(-12)^2 + 13^2}\) (dep on at least one correct component).
A1 for \(17.7\).

1. Find the gradient of line \(L\):
Rearrange \(3x - 2y = 8\) into gradient-intercept form:
\(2y = 3x - 8\)
\(y = \frac{3}{2}x - 4\)
So, the gradient of \(L\) is \(m_1 = \frac{3}{2}\).

2. Determine the gradient of the perpendicular line, \(m_2\):
\(m_2 = -\frac{1}{m_1} = -\frac{2}{3}\)

3. Find the equation of the perpendicular line through \((6, -1)\):
Using \(y - y_1 = m_2(x - x_1)\):
\(y - (-1) = -\frac{2}{3}(x - 6)\)
\(y + 1 = -\frac{2}{3}x + 4\)
\(y = -\frac{2}{3}x + 3\)

M1 for finding the gradient of the original line as \(\frac{3}{2}\) or the perpendicular gradient as \(-\frac{2}{3}\).
M1 for substituting \((6, -1)\) into \(y = \text{their } m_2 \cdot x + c\) to find \(c\).
A1 for \(y = -\frac{2}{3}x + 3\) or equivalent.

1. Convert the linear scale to a convenient unit:
\(1\text{ cm} = 25\,000\text{ cm}\)
Since \(100\text{ cm} = 1\text{ m}\) and \(1000\text{ m} = 1\text{ km}\), we have:
\(25\,000\text{ cm} = 250\text{ m} = 0.25\text{ km}\)

2. Find the area scale:
\(1\text{ cm}^2 = (0.25\text{ km})^2 = 0.0625\text{ km}^2\)

3. Calculate the actual area:
\(\text{Actual Area} = 8.4 \times 0.0625\text{ km}^2 = 0.525\text{ km}^2\)

M1 for linear scale conversion e.g. \(1\text{ cm} = 0.25\text{ km}\) or \(1\text{ cm} = 250\text{ m}\).
M1 for squaring the scale factor, e.g. \(8.4 \times 0.25^2\) or \(8.4 \times 25\,000^2 / 10^{10}\).
A1 for \(0.525\).

1. Find the fraction of the full circle represented by the sector (or the angle \(\theta\)):
\(\text{Area of sector} = \frac{\theta}{360} \times \pi r^2\)
\(13.5\pi = \frac{\theta}{360} \times \pi \times 9^2\)
\(13.5 = \frac{\theta}{360} \times 81\)
\(\frac{\theta}{360} = \frac{13.5}{81} = \frac{1}{6}\)

2. Calculate the arc length of the sector:
\(\text{Arc length} = \frac{\theta}{360} \times 2\pi r = \frac{1}{6} \times 2\pi(9) = 3\pi\text{ cm}\)

3. Calculate the total perimeter of the sector (arc length + 2 radii):
\(\text{Perimeter} = \text{Arc length} + 2r = 3\pi + 2(9) = 3\pi + 18\text{ cm}\)

M1 for setting up sector area equation to find the fraction \(\frac{1}{6}\) or angle \(\theta = 60^\circ\).
M1 for finding the arc length, \(3\pi\).
A1 for \(3\pi + 18\) or \(18 + 3\pi\).

\( \text{(a) } V_{\text{total}} = V_{\text{hemisphere}} + V_{\text{cylinder}} = \frac{2}{3}\pi r^3 + \pi r^2 h \)

\( \text{(b) } \frac{2}{3}\pi r^3 + \pi r^2 (8) = 90\pi \implies \frac{2}{3}r^3 + 8r^2 = 90 \).
Multiply the entire equation by \( \frac{3}{2} \):
\( r^3 + 12r^2 = 135 \implies r^3 + 12r^2 - 135 = 0 \).

\( \text{(c) } 3^3 + 12(3^2) - 135 = 27 + 108 - 135 = 135 - 135 = 0 \). Thus, \(r = 3\) is a solution.

\( \text{(d) } \text{Total surface area} = A_{\text{hemisphere curved}} + A_{\text{cylinder curved}} + A_{\text{cylinder base}} \)
\( = 2\pi r^2 + 2\pi r h + \pi r^2 = 3\pi r^2 + 2\pi r h \)
\( = 3\pi(3)^2 + 2\pi(3)(8) = 27\pi + 48\pi = 75\pi \text{ cm}^2 \).

\( \text{(e) } \text{Volume} = 90\pi \approx 282.743 \text{ cm}^3 \).
\( \text{Mass} = \text{Volume} \times \text{density} = 282.743 \times 8.4 = 2375.04 \text{ g} \).
Converting to kilograms: \( 2375.04 / 1000 \approx 2.375 \text{ kg} \).
To 3 significant figures, \( 2.38 \text{ kg} \).

\( \text{(a) } \)
M1 for \( \frac{2}{3}\pi r^3 \) or \( \pi r^2 h \)
A1 for \( \frac{2}{3}\pi r^3 + \pi r^2 h \)

\( \text{(b) } \)
M1 for substituting \( V = 90\pi \) and \( h = 8 \) into their formula: \( \frac{2}{3}\pi r^3 + 8
\pi r^2 = 90\pi \)
M1 for dividing by \(\pi\) and multiplying by 1.5: \( r^3 + 12r^2 = 135 \)
A1 for establishing the given result \( r^3 + 12r^2 - 135 = 0 \) clearly.

\( \text{(c) } \)
B1 for \( 3^3 + 12(3)^2 - 135 = 27 + 108 - 135 = 0 \) shown clearly.

\( \text{(d) } \)
M1 for recognizing the components: \( 2\pi r^2 \) and \( 2\pi r h \) and \( \pi r^2 \)
M1 for substituting \( r=3 \) and \( h=8 \) into their surface area expression
A1 for \( 75\pi \)

\( \text{(e) } \)
M1 for mass \( = 90\pi \times 8.4 \)
A1 for \( 2.38 \) (allow \(2.37\) to \(2.38\))

\( \text{(a) } \text{Cost} = x(x+1) + 2(x-1) = x^2 + x + 2x - 2 = x^2 + 3x - 2 \).

\( \text{(b) } 3y + 5(y + 1.50) = 29.50 \implies 3y + 5y + 7.50 = 29.50 \implies 8y = 22.00 \implies y = 2.75 \).
The cost of 1 kg of pears is \( y + 1.50 = 2.75 + 1.50 = \$4.25 \).

\( \text{(c)(i) } \text{By Pythagoras' Theorem: } w^2 + (2w + 5)^2 = 15^2 \)
\( w^2 + (4w^2 + 20w + 25) = 225 \)
\( 5w^2 + 20w + 25 - 225 = 0 \)
\( 5w^2 + 20w - 200 = 0 \) (Shown).

\( \text{(c)(ii) } \text{Dividing by 5 gives } w^2 + 4w - 40 = 0 \).
Using the quadratic formula:
\( w = \frac{-4 \pm \sqrt{4^2 - 4(1)(-40)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 160}}{2} = \frac{-4 \pm \sqrt{176}}{2} \)
\( w \approx \frac{-4 \pm 13.2665}{2} \)
\( w_1 = \frac{9.2665}{2} \approx 4.63 \)
\( w_2 = \frac{-17.2665}{2} \approx -8.63 \)

\( \text{(a) } \)
M1 for writing \( x(x+1) + 2(x-1) \)
A1 for \( x^2 + 3x - 2 \)

\( \text{(b) } \)
M1 for forming equation \( 3y + 5(y + 1.50) = 29.50 \)
M1 for solving to get \( y = 2.75 \)
A1 for \( 4.25 \)

\( \text{(c)(i) } \)
M1 for \( w^2 + (2w + 5)^2 = 15^2 \)
B1 for \( (2w + 5)^2 = 4w^2 + 20w + 25 \)
A1 for completing the algebraic proof to show \( 5w^2 + 20w - 200 = 0 \)

\( \text{(c)(ii) } \)
M1 for use of quadratic formula with correct coefficients
A1 for one correct value (e.g. \( 4.63 \) or \( -8.63 \))
A1 for both \( 4.63 \) and \( -8.63 \)

\( \text{(a) } \)
- When \( x = 1 \): \( y = \frac{12}{1} + 1^2 - 6 = 12 + 1 - 6 = 7 \)
- When \( x = 2 \): \( y = \frac{12}{2} + 2^2 - 6 = 6 + 4 - 6 = 4 \)
- When \( x = 4 \): \( y = \frac{12}{4} + 4^2 - 6 = 3 + 16 - 6 = 13 \)

\( \text{(b) } \frac{12}{x} + x^2 - 6 = 2x + 1 \)
Multiply all terms by \( x \):
\( 12 + x^3 - 6x = x(2x + 1) \)
\( 12 + x^3 - 6x = 2x^2 + x \)
Rearrange all terms to one side:
\( x^3 - 2x^2 - 6x - x + 12 = 0 \)
\( x^3 - 2x^2 - 7x + 12 = 0 \) (Shown).

\( \text{(c)(i) } y = 12x^{-1} + x^2 - 6 \)
Differentiating with respect to \( x \):
\( \frac{\text{dy}}{\text{dx}} = -12x^{-2} + 2x = -\frac{12}{x^2} + 2x \)
At \( x = 2 \):
\( \frac{\text{dy}}{\text{dx}} = -\frac{12}{2^2} + 2(2) = -\frac{12}{4} + 4 = -3 + 4 = 1 \).
So the gradient \( m = 1 \).

\( \text{(c)(ii) } \text{Since the tangent passes through the point } (2, 4) \text{ and has gradient } m = 1 \text{:} \)
\( y = mx + c \implies 4 = 1(2) + c \implies c = 2 \).

\( \text{(a) } \)
B1 for \( y = 7 \) when \( x = 1 \)
B1 for \( y = 4 \) when \( x = 2 \)
B1 for \( y = 13 \) when \( x = 4 \)

\( \text{(b) } \)
M1 for multiplying by \( x \) to get \( 12 + x^3 - 6x = x(2x + 1) \)
M1 for expanding \( x(2x + 1) \) to get \( 2x^2 + x \)
A1 for completing the algebra to get \( x^3 - 2x^2 - 7x + 12 = 0 \) with no errors.

\( \text{(c)(i) } \)
M1 for differentiating to find at least one correct term: \( -12x^{-2} \) or \( 2x \)
A1 for the complete derivative \( \frac{\text{dy}}{\text{dx}} = -\frac{12}{x^2} + 2x \)
A1 for substituting \( x = 2 \) to get \( 1 \) clearly.

\( \text{(c)(ii) } \)
M1 for substituting their \( (2, 4) \) and \( m = 1 \) into \( y = mx + c \)
A1 for \( c = 2 \)

\( \text{(a) } \text{Using the Cosine Rule:} \)
\( PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(\angle PQR) \)
\( PR^2 = 85^2 + 120^2 - 2(85)(120)\cos(72^\circ) \)
\( PR^2 = 7225 + 14400 - 20400 \cdot 0.309017 \)
\( PR^2 = 21625 - 6303.95 = 15321.05 \)
\( PR = \sqrt{15321.05} \approx 123.778 \text{ m} \approx 123.8 \text{ m} \).

\( \text{(b) } \text{Area} = \frac{1}{2} \cdot PQ
\cdot QR \cdot \sin(\angle PQR) \)
\( \text{Area} = \frac{1}{2}(85)(120)\sin(72^\circ) \approx 5100 \cdot 0.9510565 = 4850.39 \text{ m}^2 \).
To the nearest square metre, \( \text{Area} = 4850 \text{ m}^2 \).

\( \text{(c) } \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot PR \cdot QS \)
\( 4850.39 = \frac{1}{2}(123.778) \cdot QS \)
\( QS = \frac{2
\cdot 4850.39}{123.778} \approx 78.37 \text{ m} \approx 78.4 \text{ m} \).

\( \text{(d) } \text{Using the Sine Rule:} \)
\( \frac{\sin(\angle QRP)}{PQ} = \frac{\sin(\angle PQR)}{PR} \)
\( \frac{\sin(\angle QRP)}{85} = \frac{\sin(72^\circ)}{123.778} \)
\(
\sin(\angle QRP) = \frac{85 \sin(72^\circ)}{123.778} \approx 0.65315 \)
\( \angle QRP = \sin^{-1}(0.65315) \approx 40.78^\circ \approx 40.8^\circ \).

\( \text{(a) } \)
M1 for \( 85^2 + 120^2 - 2(85)(120)\cos(72) \)
A1 for \( 15321.05 \)
A1 for \( 123.8 \) (allow \(123.7\) to \(123.8\))

\( \text{(b) } \)
M1 for \( \frac{1}{2}(85)(120)\sin(72) \)
A1 for \( 4850 \)

\( \text{(c) } \)
M1 for equating area formula: \( \frac{1}{2} \times \text{their } PR
\times QS = \text{their Area} \)
M1 for \( QS = \frac{2 \times \text{their Area}}{\text{their } PR} \)
A1 for \( 78.4 \) (allow \(78.3\) to \(78.5\))

\( \text{(d) } \)
M1 for \( \frac{\sin(\angle QRP)}{85} = \frac{\sin(72)}{123.8} \) or equivalent Cosine Rule setup
M1 for \( \sin(\angle QRP) \approx 0.653 \) or \(
\cos(\angle QRP) \approx 0.758 \)
A1 for \( 40.8^\circ \) (allow \(40.7^\circ\) to \(40.9^\circ\))

\( \text{(a)(i) } \text{The ratio of investments is } 3 : 4 : 5 \text{ (Arthur : Beatrice : Charlie).} \)
\( \text{Beatrice's investment represents } 4 \text{ parts of the total } 3 + 4 + 5 = 12 \text{ parts.} \)
\( 1 \text{ part} = \frac{\$18,000}{4} = \$4,500 \).
\( \text{Total investment} = 12 \text{ parts} = 12 \times \$4,500 = \$54,000 \). (Shown).

\( \text{(a)(ii) } \text{Charlie's investment} = 5 \text{ parts} = 5 \times \$4,500 = \$22,500 \).

\( \text{(b) } \text{Total profit} = 14\% \text{ of } \$54,000 = 0.14 \times 54,000 = \$7,560 \).
\( \text{Arthur's share is } \frac{3}{12} = \frac{1}{4} \text{ of the total profit.} \)
\( \text{Arthur's profit} = \frac{1}{4} \times \$7,560 = \$1,890 \).

\( \text{(c) } \text{Let the original investment be } I \text{.} \)
\( \text{After year 2, value} = I
\times (1 - 0.08) = 0.92I \).
\( \text{After year 3, value} = 0.92I \times (1 + 0.12) = 0.92 \times 1.12I = 1.0304I \).
\( \text{The overall change is a factor of } 1.0304 \text{, which is an increase of } (1.0304 - 1) \times 100\% = 3.04\% \).

\( \text{(d) } 13500 \times \left(1 + \frac{r}{100}\right)^4 = 15000 \)
\( \left(1 + \frac{r}{100}\right)^4 = \frac{15000}{13500} = \frac{10}{9} \)
\( 1 + \frac{r}{100} = \left(\frac{10}{9}\right)^{0.25} \approx 1.026684 \)
\( \frac{r}{100} \approx 0.026684 \implies r \approx 2.67 \).

\( \text{(a)(i) } \)
M1 for \( 18000 \div 4 \) or establishing that 1 part is \( 4500 \)
A1 for completing the calculation \( 12 \times 4500 = 54000 \) clearly.

\( \text{(a)(ii) } \)
B1 for \( 22500 \)

\( \text{(b) } \)
M1 for \( 0.14 \times 54000 \) or \( 7560 \)
M1 for \(
\frac{3}{12}
\times (\text{their total profit}) \)
A1 for \( 1890 \)

\( \text{(c) } \)
M1 for multiplier \( 0.92 \) or \( 1.12 \)
M1 for \( 0.92 \times 1.12 = 1.0304 \)
A1 for \( 3.04\% \) increase (must state 'increase' or use \(+\))

\( \text{(d) } \)
M1 for \( 13500(1 + \frac{r}{100})^4 = 15000 \) or \( (1 + \frac{r}{100}) = \sqrt[4]{\frac{15000}{13500}} \)
A1 for \( r = 2.67 \) (allow \(2.66\) to \(2.68\))

\( \text{(a) } \)
- For \( t \le 200 \): Cumulative Frequency \( = 15 + 28 = 43 \)
- For \( t \le 250 \): Cumulative Frequency \( = 43 + 42 = 85 \)
- For \( t \le 300 \): Cumulative Frequency \( = 85 + 25 = 110 \)

\( \text{(b)(i) } \text{The median occurs at Cumulative Frequency } = 60 \text{.} \)
\( \text{The interval containing 60 is } 200 < t \le 250 \text{, between the points } (200, 43) \text{ and } (250, 85) \text{.} \)
\( \text{Using linear interpolation:} \)
\( \text{Median} = 200 +
\frac{60 - 43}{85 - 43} \times (250 - 200) = 200 + \frac{17}{42} \times 50 \approx 200 + 20.24 = 220.2 \text{ hours} \).

\( \text{(b)(ii) } \text{The lower quartile (LQ) occurs at Cumulative Frequency } = 30 \text{.} \)
\( \text{The interval containing 30 is } 150 < t \le 200 \text{, between the points } (150, 15) \text{ and } (200, 43) \text{.} \)
\( \text{LQ} = 150 +
\frac{30 - 15}{43 - 15} \times (200 - 150) = 150 + \frac{15}{28} \times 50 \approx 150 + 26.79 = 176.79 \text{ hours} \).
\( \text{The upper quartile (UQ) occurs at Cumulative Frequency } = 90 \text{.} \)
\( \text{The interval containing 90 is } 250 < t \le 300 \text{, between the points } (250, 85) \text{ and } (300, 110) \text{.} \)
\( \text{UQ} = 250 +
\frac{90 - 85}{110 - 85} \times (300 - 250) = 250 + \frac{5}{25} \times 50 = 250 + 10 = 260 \text{ hours} \).
\( \text{IQR} =
\text{UQ} - \text{LQ} = 260 - 176.79 = 83.21 \approx 83.2 \text{ hours} \).

\( \text{(c) } \text{To find the cumulative frequency at } t = 280 \text{ hours (which lies in } 250 < t \le 300\text{):} \)
\( \text{Cumulative Frequency} = 85 +
\frac{280 - 250}{300 - 250} \times (110 - 85) = 85 + \frac{30}{50} \times 25 = 85 + 15 = 100 \).
This means 100 light bulbs have a lifetime of 280 hours or less.
Therefore, the number of light bulbs with a lifetime of more than 280 hours is:
\( 120 - 100 = 20 \).

\( \text{(a) } \)
B1 for two correct values (e.g. 43 and 85)
B1 for third correct value (110)

\( \text{(b)(i) } \)
M1 for realizing the median is at CF \( = 60 \) and setting up the fraction \( \frac{60 - 43}{85 - 43} \) or equivalent graph-reading method.
A1 for \( 220.2 \) (allow \(220\) to \(221\))

\( \text{(b)(ii) } \)
M1 for LQ at CF \( = 30 \) and UQ at CF \( = 90 \)
M1 for LQ \( \approx 176.8 \)
M1 for UQ \( = 260 \)
A1 for \( 83.2 \) (allow \(83.0\) to \(83.5\))

\( \text{(c) } \)
M1 for locating \( t = 280 \) in the interval \( 250 < t \le 300 \)
M1 for finding CF \( = 100 \) at \( t = 280 \)
A1 for \( 20 \)

\( \text{(a)(i) } \text{Sequence A is arithmetic with a common difference of } 3 \text{. The next term is } 17 + 3 = 20 \text{.} \)
\( \text{(a)(ii) } \text{Sequence B has differences } 4, 6, 8, 10, \dots \text{ The next difference is } 12 \text{, so the next term is } 30 + 12 = 42 \text{.} \)
\( \text{(a)(iii) } \text{Sequence C is geometric with a common ratio of } 2 \text{. The next term is } 64 \times 2 = 128 \text{.} \)

\( \text{(b)(i) } \text{Arithmetic sequence with } a = 5 \text{ and } d = 3 \text{:} \)
\( u_n = a + (n-1)d = 5 + (n-1)(3) = 3n + 2 \).

\( \text{(b)(ii) } \text{Sequence B: } 2, 6, 12, 20, 30 \).
First differences: \( 4, 6, 8, 10 \)
Second differences: \( 2, 2, 2 \)
Since the second difference is constant and equals 2, the \( n^2 \) coefficient is \( \frac{2}{2} = 1 \).
Subtracting \( n^2 \) from each term of the sequence:
- \( 2 - 1^2 = 1 \)
- \( 6 - 2^2 = 2 \)
- \( 12 - 3^2 = 3 \)
- \( 20 - 4^2 = 4 \)
This gives the linear sequence \( 1, 2, 3, 4, \dots \), which has \( n \)-th term \( n \).
Thus, the \( n \)-th term of Sequence B is \( n^2 + n \).

\( \text{(b)(iii) } \text{Sequence C is } 4, 8, 16, 32, 64 \dots \) which can be written as \( 2^2, 2^3, 2^4, 2^5, 2^6 \dots \)
The power of 2 is \( n + 1 \).
Thus, the \( n \)-th term is \( 2^{n+1} \).

\( \text{(c) } n^2 + n = n(n + 1) \). Since \( n \) and \( n + 1 \) are consecutive integers, \( n^2 + n \) is indeed the product of two consecutive integers.

\( \text{(a)(i) } \)
B1 for \( 20 \)

\( \text{(a)(ii) } \)
B1 for \( 42 \)

\( \text{(a)(iii) } \)
B1 for \( 128 \)

\( \text{(b)(i) } \)
M1 for a linear expression of the form \( 3n + k \) where \( k \ is a constant.
A1 for \) 3n + 2 \)

\( \text{(b)(ii) } \)
M1 for recognizing second differences are constant and writing \( a n^2 + b n + c \) with \( a = 1 \).
M1 for trying to find \( b \) and \( c \) by setting up equations or subtracting \( n^2 \).
A1 for \( n^2 + n \)

\( \text{(b)(iii) } \)
M1 for recognizing powers of 2 (e.g., \( 2^k \))
A1 for \( 2^{n+1} \) (or \( 4 \times 2^{n-1} \))

\( \text{(c) } \)
B1 for factoring to \( n(n+1) \) and stating that \( n \) and \( n+1 \) are consecutive integers.

\( \text{(a) } \angle ACD = 180^\circ - \angle DCT = 180^\circ - 124^\circ = 56^\circ \) because angles on a straight line (\(ACT\)) sum to \(180^\circ\).

\( \text{(b) } \angle ADC = 90^\circ \) because the angle subtended by a diameter (\(AC\)) in a semicircle is a right angle.

\( \text{(c) } \text{In triangle } ADC\text{, the angles sum to } 180^\circ\text{.} \)
\( \angle CAD = 180^\circ - \angle ADC - \angle ACD = 180^\circ - 90^\circ - 56^\circ = 34^\circ \).

\( \text{(d) } \text{By the Alternate Segment Theorem, the angle between the tangent } TD \text{ and the chord } CD \text{ is equal to the angle subtended by } CD \text{ in the alternate segment, which is } \angle CAD\text{.} \)
\( \text{Therefore, } \angle CDT = \angle CAD = 34^\circ \).
\( \text{Alternatively, } OD \perp TD \implies \angle ODT = 90^\circ \text{. Since } OD = OC \text{ (radii), } \angle ODC = \angle OCD = \angle ACD = 56^\circ \text{.} \)
\( \angle CDT = \angle ODT - \angle ODC = 90^\circ - 56^\circ = 34^\circ \).

\( \text{(e) } \text{In triangle } DCT\text{, the angles sum to } 180^\circ\text{.} \)
\( \angle DTA = \angle DTC = 180^\circ - \angle DCT - \angle CDT = 180^\circ - 124^\circ - 34^\circ = 22^\circ \).

\( \text{(a) } \)
M1 for \( 180 - 124 \)
A1 for \( 56^\circ \) with valid reason (angles on a straight line)

\( \text{(b) } \)
B1 for \( 90^\circ \)
B1 for valid reason (angle in a semicircle / angle subtended by diameter)

\( \text{(c) } \)
M1 for \( 180 - 90 - 56 \) (using their answers to a and b)
A1 for \( 34^\circ \)

\( \text{(d) } \)
M1 for identifying the relevant circle theorem (Alternate Segment Theorem) or using radius-tangent theorem.
M1 for \( \angle ODC = 56^\circ \) or stating \(
\angle CDT = \angle CAD \)
A1 for \( 34^\circ \)

\( \text{(e) } \)
M1 for \( 180 - 124 - 34 \) (using their answer to d)
A1 for \( 22^\circ \)

(a) Using the Cosine Rule to find \(AC\):
\(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)\)
\(AC^2 = 120^2 + 150^2 - 2 \times 120 \times 150 \times \cos(78^\circ)\)
\(AC^2 = 14400 + 22500 - 36000 \times 0.20791 = 29415.18\)
\(AC = \sqrt{29415.18} \approx 171.51\text{ m}\)

(b) Using the Sine Rule to find angle \(BAC\):
\(\frac{\sin(BAC)}{BC} = \frac{\sin(ABC)}{AC}\)
\(\frac{\sin(BAC)}{150} = \frac{\sin(78^\circ)}{171.51}\)
\(\sin(BAC) = \frac{150 \times \sin(78^\circ)}{171.51} \approx 0.85547\)
\(\text{Angle } BAC = \sin^{-1}(0.85547) \approx 58.81^\circ\)

(c) Calculating the area of triangle \(ABC\):
\(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(ABC)\)
\(\text{Area} = \frac{1}{2} \times 120 \times 150 \times \sin(78^\circ) = 9000 \times 0.97815 = 8803.35\text{ m}^2\)

(d) The shortest distance from \(B\) to \(AC\) is the perpendicular height, \(d\):
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times d\)
\(8803.35 = \frac{1}{2} \times 171.51 \times d\)
\(d = \frac{2 \times 8803.35}{171.51} \approx 102.66\text{ m}\)

(a) M1 for correct substitution into Cosine Rule: \(120^2 + 150^2 - 2 \times 120 \times 150 \times \cos(78)\)
A1 for 29415
A1 for 171.5 [accept 171.5 to 172]

(b) M1 for implicit Sine Rule: \(\frac{\sin(BAC)}{150} = \frac{\sin(78)}{\text{their (a)}}\)
M1 for explicit rearrangement: \(\sin(BAC) = \frac{150 \times \sin(78)}{\text{their (a)}}\)
A1 for 0.855...
A1 for 58.8 [accept 58.7 to 58.9]

(c) M1 for \(0.5 \times 120 \times 150 \times \sin(78)\)
A1 for 8800 or 8803 [accept 8800 to 8810]

(d) M1 for \(120 \times \sin(\text{their } BAC)\) or \(0.5 \times \text{their (a)} \times d = \text{their (c)}\)
A1 for 103 or 102.7 [accept 102.5 to 103]

(a) First, find the vertical height of the cone, \(h_{\text{cone}}\):
\(h_{\text{cone}} = \sqrt{4.5^2 - 3.2^2} = \sqrt{20.25 - 10.24} = \sqrt{10.01} \approx 3.164\text{ m}\)

\(\text{Volume of cylinder} = \pi r^2 h = \pi \times 3.2^2 \times 8.5 = 87.04\pi \approx 273.45\text{ m}^3\)
\(\text{Volume of cone} = \frac{1}{3} \pi r^2 h_{\text{cone}} = \frac{1}{3} \pi \times 3.2^2 \times 3.164 \approx 33.93\text{ m}^3\)
\(\text{Total volume} = 273.45 + 33.93 = 307.38\text{ m}^3\)

(b) \(\text{Curved surface area of cylinder} = 2\pi r h = 2 \times \pi \times 3.2 \times 8.5 = 54.4\pi \approx 170.90\text{ m}^2\)
\(\text{Curved surface area of cone} = \pi r l = \pi \times 3.2 \times 4.5 = 14.4\pi \approx 45.24\text{ m}^2\)
\(\text{Total external surface area} = 170.90 + 45.24 = 216.14\text{ m}^2\)

(c) Total height of the original silo: \(H_{\text{silo}} = 8.5 + 3.164 = 11.664\text{ m}\).
Volume ratio: \(\frac{V_{\text{model}}}{V_{\text{original}}} = \frac{1.5}{307.38} \approx 0.004880\)
Linear scale factor \(k = \sqrt[3]{0.004880} \approx 0.1696\)
Height of the model = \(11.664 \times 0.1696 \approx 1.978\text{ m} = 197.8\text{ cm}\)

(a) M1 for finding vertical height of cone: \(\sqrt{4.5^2 - 3.2^2}\) or 3.16
M1 for cylinder volume: \(\pi \times 3.2^2 \times 8.5\) (273.4 to 273.5)
M1 for cone volume: \(\frac{1}{3} \pi \times 3.2^2 \times 3.16\) (33.9 to 34.0)
A1 for 307 or 307.4 [accept 307 to 308]

(b) M1 for cylinder curved surface: \(2 \pi \times 3.2 \times 8.5\) (170.9)
M2 for cone curved surface: \(\pi \times 3.2 \times 4.5\) (45.2)
A1 for 216 [accept 216 to 216.3]

(c) M1 for adding heights of original silo to get 11.66 or 11.7
M1 for \(\sqrt[3]{\frac{1.5}{\text{their (a)}}} \times \text{their } H_{\text{original}}\)
A1 for 198 or 197.8 [accept 197 to 199]

(a) For \(x = -2\): \(y = (-2)^3 - 4(-2) + 2 = -8 + 8 + 2 = 2 \implies p = 2\)
For \(x = 1\): \(y = 1^3 - 4(1) + 2 = -1 \implies q = -1\)
For \(x = 3\): \(y = 3^3 - 4(3) + 2 = 27 - 12 + 2 = 17 \implies r = 17\)

(b) Drawing \(y = 2x - 1\) (e.g., passes through \((0, -1)\) and \((2, 3)\)).
The equation \(x^3 - 4x + 2 = 2x - 1\) simplifies to \(x^3 - 6x + 3 = 0\).
The roots of this equation are the \(x\)-coordinates of the intersection points:
\(x \approx -2.63\), \(x \approx 0.54\), and \(x \approx 2.09\).

(c) Draw a tangent line to the curve at the point \((2, 2)\).
Using two points on the tangent line, e.g., \((1, -6)\) and \((3, 10)\):
\(\text{Gradient} = \frac{10 - (-6)}{3 - 1} = \frac{16}{2} = 8\).

(a) B1 for \(p = 2\)
B1 for \(q = -1\)
B1 for \(r = 17\)

(b) M1 for drawing a straight line through \((0, -1)\) with gradient 2.
A1 for one correct root [accept: \(-2.7\) to \(-2.5\)]
A1 for second correct root [accept: \(0.4\) to \(0.7\)]
A1 for third correct root [accept: \(2.0\) to \(2.2\)]

(c) M1 for drawing a tangent to the curve at \(x = 2\) (must touch the curve at \(x = 2\) only)
M2 for finding gradient from their tangent: \(\frac{y_2 - y_1}{x_2 - x_1}\) with clear working shown
A1 for answer in range \([7.0, 9.0]\)

(a) \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\)
\(\text{Total time} = \frac{36}{x} + \frac{24}{x - 3}\) hours.

(b) Convert \(3\text{ hours } 40\text{ minutes}\) to hours: \(3\frac{40}{60} = 3\frac{2}{3} = \frac{11}{3}\) hours.
Set up the equation: \(\frac{36}{x} + \frac{24}{x - 3} = \frac{11}{3}\)
Multiply the entire equation by \(3x(x-3)\) to clear denominators:
\(3 \times 36(x - 3) + 3 \times 24x = 11x(x - 3)\)
\(108(x - 3) + 72x = 11x^2 - 33x\)
\(108x - 324 + 72x = 11x^2 - 33x\)
\(180x - 324 = 11x^2 - 33x\)
\(11x^2 - 213x + 324 = 0\).

(c) Use the quadratic formula with \(a = 11\), \(b = -213\), \(c = 324\):
\(x = \frac{-(-213) \pm \sqrt{(-213)^2 - 4(11)(324)}}{2 \times 11}\)
\(x = \frac{213 \pm \sqrt{45369 - 14256}}{22}\)
\(x = \frac{213 \pm \sqrt{31113}}{22}\)
\(x = \frac{213 \pm 176.38878}{22}\)
\(x_1 = \frac{389.38878}{22} \approx 17.70\)
\(x_2 = \frac{36.61122}{22} \approx 1.66\)

(d) If \(x = 1.66\), then the speed for the second part is \(x - 3 = 1.66 - 3 = -1.34\text{ km/h}\). A speed cannot be negative.

(a) M1 for \(\frac{36}{x}\) or \(\frac{24}{x-3}\)
A1 for \(\frac{36}{x} + \frac{24}{x - 3}\)

(b) M1 for \(3\text{h } 40\text{m} = \frac{11}{3}\) hours
M1 for expanding with common denominator: \(36(x-3) + 24x = \frac{11}{3}x(x-3)\) or equivalent
M1 for multiplying out brackets: \(108x - 324 + 72x = 11x^2 - 33x\)
A1 for fully correct simplification leading to \(11x^2 - 213x + 324 = 0\)

(c) M1 for correct substitution into formula: \(\frac{-(-213) \pm \sqrt{(-213)^2 - 4(11)(324)}}{2 \times 11}\)
B1 for \(\sqrt{31113}\) or \(176.4...\)
A1 for 17.70
A1 for 1.66

(d) B1 for stating that \(x - 3\) would be negative / speed must be positive