2025 Edexcel IGCSE Mathematics (Specification A) 模擬試題連答案詳解

Express both sides of the equation as powers of 2:
\( 8 = 2^3 \implies 8^{2x-1} = (2^3)^{2x-1} = 2^{3(2x-1)} = 2^{6x-3} \)
\( \frac{1}{32} = 32^{-1} = (2^5)^{-1} = 2^{-5} \)

Set the indices equal to each other:
\( 6x - 3 = -5 \)
\( 6x = -2 \)
\( x = -\frac{2}{6} = -\frac{1}{3} \)

M1 for expressing both sides as powers of 2, e.g., \( 2^{3(2x-1)} \) or \( 2^{6x-3} \) and \( 2^{-5} \)
M1 for setting up the linear equation \( 6x - 3 = -5 \) and attempting to solve for \( x \)
A1 for \( -\frac{1}{3} \) (or equivalent fraction or recurring decimal)

First, factorise the numerator:
\( 2x^2 + 5x - 3 = (2x - 1)(x + 3) \)

Next, factorise the denominator using the difference of two squares:
\( x^2 - 9 = (x - 3)(x + 3) \)

Now, simplify by cancelling the common factor \( (x + 3) \):
\( \frac{(2x - 1)(x + 3)}{(x - 3)(x + 3)} = \frac{2x - 1}{x - 3} \)

M1 for factorising the numerator correctly as \( (2x - 1)(x + 3) \) or denominator correctly as \( (x - 3)(x + 3) \)
M1 for factorising both correctly
A1 for \( \frac{2x-1}{x-3} \)

The probability of taking at least one green sweet is equal to \( 1 - \text{P(no green sweets)} \).
No green sweets means both sweets taken are red.

Probability of first sweet being red is \( \frac{4}{9} \).
Since there is no replacement, there are now 3 red sweets and 8 sweets in total.
Probability of second sweet being red is \( \frac{3}{8} \).

\( \text{P(both red)} = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72} = \frac{1}{6} \)

Therefore, \( \text{P(at least one green)} = 1 - \frac{1}{6} = \frac{5}{6} \).

M1 for finding the probability of drawing two red sweets, e.g., \( \frac{4}{9} \times \frac{3}{8} \)
M1 for completing the calculation \( 1 - \text{P(both red)} \)
A1 for \( \frac{5}{6} \) (or equivalent decimal \( 0.833... \) or \( 83.3\% \))

Let the total number of parts in the ratio be \( 3 + 7 = 10 \).
We can assume there are 30 part-time workers and 70 full-time workers (a total of 100 workers).

Number of male part-time workers = \( 40\% \text{ of } 30 = 0.4 \times 30 = 12 \).
Number of male full-time workers = \( 30\% \text{ of } 70 = 0.3 \times 70 = 21 \).

Total number of male workers = \( 12 + 21 = 33 \).
Total number of workers = 100.

Fraction of male workers = \( \frac{33}{100} \).
This fraction cannot be simplified further as 33 and 100 share no common factors.

M1 for a correct method to find the proportion/number of male workers in either group, e.g., \( 0.4 \times 3 \) or \( 0.3 \times 7 \)
M1 for adding the two proportions/numbers to find the total male representation, e.g., \( 1.2 + 2.1 = 3.3 \) out of 10
A1 for \( \frac{33}{100} \) (must be in simplest form)

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

Substitute the given values:
\( AC^2 = 8^2 + 11^2 - 2(8)(11)\cos(120^\circ) \)
\( AC^2 = 64 + 121 - 176(-0.5) \)
\( AC^2 = 185 + 88 = 273 \)

Calculate \( AC \):
\( AC = \sqrt{273} \approx 16.5227\text{ cm} \)

Correct to 3 significant figures, the length is \( 16.5\text{ cm} \).

M1 for substituting the given values into the Cosine Rule formula, e.g., \( 8^2 + 11^2 - 2 \times 8 \times 11 \times \cos(120^\circ) \)
M1 for evaluating to \( 273 \) or showing \( \sqrt{273} \)
A1 for \( 16.5 \) (accept 16.5 to 16.53)

The volume of a hemisphere is given by:
\( V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 \)

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

Since \( h = 4r \):
\( V_{\text{cone}} = \frac{1}{3}\pi r^2 (4r) = \frac{4}{3}\pi r^3 \)

The total volume of the solid is:
\( V_{\text{total}} = \frac{2}{3}\pi r^3 + \frac{4}{3}\pi r^3 = 2\pi r^3 \)

We are given \( V_{\text{total}} = 250\pi \):
\( 2\pi r^3 = 250\pi \)
\( r^3 = 125 \)
\( r = 5 \)

M1 for expressing the volume of either the hemisphere or the cone in terms of \( r \), e.g., \( \frac{2}{3}\pi r^3 \) or \( \frac{1}{3}\pi r^2(4r) \)
M1 for equating the sum of both volumes to \( 250\pi \), e.g., \( 2\pi r^3 = 250\pi \) or \( r^3 = 125 \)
A1 for \( 5 \)

First, find the composite function \( fg(x) \):
\( fg(x) = f(g(x)) = f(3x+1) = \frac{4}{(3x+1) - 2} = \frac{4}{3x - 1} \)

Now, set \( fg(x) = 2 \) and solve for \( x \):
\( \frac{4}{3x - 1} = 2 \)
\( 4 = 2(3x - 1) \)
\( 4 = 6x - 2 \)
\( 6x = 6 \)
\( x = 1 \)

M1 for writing an expression for \( fg(x) \), e.g., \( \frac{4}{3x+1-2} \) or \( \frac{4}{3x-1} \)
M1 for setting up the equation \( \frac{4}{3x-1} = 2 \) and attempting to solve for \( x \)
A1 for \( 1 \)

First, find the derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{d}{dx}(2x^3 - 9x^2 - 24x + 7) = 6x^2 - 18x - 24 \)

At stationary points, the gradient is zero (\( \frac{dy}{dx} = 0 \)):
\( 6x^2 - 18x - 24 = 0 \)

Divide the equation by 6 to simplify:
\( x^2 - 3x - 4 = 0 \)

Factorise the quadratic equation:
\( (x - 4)(x + 1) = 0 \)

This gives:
\( x = 4 \) or \( x = -1 \)

M1 for differentiating the equation correctly to find \( \frac{dy}{dx} = 6x^2 - 18x - 24 \) (at least two terms correct)
M1 for setting \( \frac{dy}{dx} = 0 \) and attempting to solve the quadratic equation
A1 for \( -1, 4 \) (or \( 4, -1 \))

First, write the terms in the numerator to have the same power of 10:
\(1.4 \times 10^5 = 14 \times 10^4\)

Now add the terms in the numerator:
\(14 \times 10^4 + 9.8 \times 10^4 = 23.8 \times 10^4 = 2.38 \times 10^5\)

Next, divide by the denominator:
\(\frac{2.38 \times 10^5}{3.5 \times 10^{-2}} = \left(\frac{2.38}{3.5}\right) \times 10^{5 - (-2)}\)

Calculate the division of the coefficients:
\(\frac{2.38}{3.5} = 0.68\)

Combine and express the final result in standard form:
\(0.68 \times 10^7 = 6.8 \times 10^6\)

\textbf{M1}: For converting the numerator to a single value, e.g., \(2.38 \times 10^5\) or \(238000\), or for demonstrating correct division of coefficients.
\textbf{A1.8}: For \(6.8 \times 10^6\) (accept equivalent standard form expression).

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

List the elements of set \(A\) that are in \(\mathcal{E}\) (factors of 24):
\(A = \{1, 2, 3, 4, 6, 8, 12\}\)

List the elements of set \(B\) that are in \(\mathcal{E}\) (multiples of 3):
\(B = \{3, 6, 9, 12\}\)

Find the union of sets \(A\) and \(B\):
\(A \cup B = \{1, 2, 3, 4, 6, 8, 9, 12\}\)

Find the complement of \(A \cup B\) (elements in \(\mathcal{E}\) but not in \(A \cup B\)):
\((A \cup B)' = \{5, 7, 10, 11\}\)

\textbf{M1}: For listing the elements of \(A \cup B\) correctly, or for identifying the correct elements of both set \(A\) and set \(B\).
\textbf{A1.8}: For the correct list of elements \(\{5, 7, 10, 11\}\) (order does not matter, brackets are not strictly required).

To subtract the fractions, find a common denominator, which is \((2x-1)(3x+2)\):

\(\frac{3(3x+2) - 4(2x-1)}{(2x-1)(3x+2)}\)

Expand the terms in the numerator:
\(3(3x+2) = 9x + 6\)
\(-4(2x-1) = -8x + 4\)

Combine like terms:
\(9x + 6 - 8x + 4 = x + 10\)

This gives the fully simplified fraction:
\(\frac{x+10}{(2x-1)(3x+2)}\)

\textbf{M1}: For writing the fractions over a common denominator with at least one numerator expanded correctly.
\textbf{A1.8}: For the fully simplified fraction \(\frac{x+10}{(2x-1)(3x+2)}\) or \(\frac{x+10}{6x^2+x-2}\).

Set up the equation:
\(3n^2 - 5n = 250\)

Rearrange into a quadratic equation equal to 0:
\(3n^2 - 5n - 250 = 0\)

Solve the quadratic equation using the quadratic formula or factorisation:
\(n = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-250)}}{2(3)}
= \frac{5 \pm \sqrt{25 + 3000}}{6}
= \frac{5 \pm \sqrt{3025}}{6}
= \frac{5 \pm 55}{6}\)

This yields two potential solutions:
\(n = \frac{60}{6} = 10\) or \(n = \frac{-50}{6} = -8.33\)

Since the term number \(n\) must be a positive integer, we select \(n = 10\).

\textbf{M1}: For setting up the quadratic equation \(3n^2 - 5n - 250 = 0\) or showing a valid method to solve it.
\textbf{A1.8}: For \(10\) as the final answer (rejecting the negative solution).

Multiply all terms in the equation by 12 (the lowest common multiple of 4 and 3) to clear the fractions:
\(12 \times \left(\frac{3x - 1}{4}\right) - 12 \times \left(\frac{x + 2}{3}\right) = 12 \times 2\)

This simplifies to:
\(3(3x - 1) - 4(x + 2) = 24\)

Expand the brackets:
\(9x - 3 - 4x - 8 = 24\)

Collect like terms:
\(5x - 11 = 24\)

Add 11 to both sides:
\(5x = 35\)

Divide by 5:
\(x = 7\)

\textbf{M1}: For multiplying by 12 (or a common denominator) and expanding brackets with at most one sign error.
\textbf{A1.8}: For the correct solution \(x = 7\).

Let \(x\) be the normal price of the laptop.
Since the price is reduced by 15%, the sale price represents 85% of the original price.

\(0.85x = 544\)

Solve for \(x\):
\(x = \frac{544}{0.85}\)
\(x = 640\)

Therefore, the normal price is £640.

\textbf{M1}: For a complete method to find the original price, e.g., \(\frac{544}{0.85}\) or showing that \(85\% = 544\).
\textbf{A1.8}: For the correct price of \(640\) (or £640).

First, expand the left-hand side of the inequality:
\(3x - 12 < 7x + 8\)

Rearrange the inequality to group the \(x\) terms on one side:
\(-12 - 8 < 7x - 3x\)
\(-20 < 4x\)

Divide by 4:
\(-5 < x\) or \(x > -5\)

The smallest integer that is strictly greater than \(-5\) is \(-4\).

\textbf{M1}: For expanding the brackets and correctly rearranging the inequality to find \(x > -5\).
\textbf{A1.8}: For identifying \(-4\) as the smallest integer satisfying the inequality.

Apply the negative power by taking the reciprocal of the fraction:
\(\left(\frac{8w^9}{27y^6}\right)^{-\frac{2}{3}} = \left(\frac{27y^6}{8w^9}\right)^{\frac{2}{3}}\)

Next, apply the fractional index \(\frac{2}{3}\) by first finding the cube root of the numerator and denominator:
\(\left(\frac{27y^6}{8w^9}\right)^{\frac{1}{3}} = \frac{\sqrt[3]{27} \cdot y^{6 \times \frac{1}{3}}}{\sqrt[3]{8} \cdot w^{9 \times \frac{1}{3}}} = \frac{3y^2}{2w^3}\)

Now, square the result:
\(\left(\frac{3y^2}{2w^3}\right)^2 = \frac{9y^4}{4w^6}\)

\textbf{M1}: For taking the reciprocal or finding the cube root of the expression correctly, e.g., showing \(\frac{3y^2}{2w^3}\) or \(\left(\frac{27y^6}{8w^9}\right)^{\frac{2}{3}}\).
\textbf{A1.8}: For the correct final simplified expression \(\frac{9y^4}{4w^6}\) (or equivalent index form such as \(2.25y^4w^{-6}\)).

Multiply both sides by \((5 + p)\):
\(w(5 + p) = 4p - 3\)

Expand the brackets:
\(5w + wp = 4p - 3\)

Rearrange to get all terms with \(p\) on one side:
\(5w + 3 = 4p - wp\)

Factorise \(p\) on the right-hand side:
\(5w + 3 = p(4 - w)\)

Divide by \((4 - w)\):
\(p = \frac{5w + 3}{4 - w}\)

(Note: \(p = \frac{-5w - 3}{w - 4}\) is also an acceptable form.)

M1: for multiplying both sides by \(5 + p\) and expanding correctly: \(5w + wp = 4p - 3\) (or equivalent)
M1: for isolating terms in \(p\) on one side and factorising: \(p(4 - w) = 5w + 3\) (or equivalent)
A1: for \(p = \frac{5w + 3}{4 - w}\) (or equivalent)

First, write the terms in the numerator with the same power of 10:
\(4.2 \times 10^7 = 4.2 \times 10^7\)
\(3.8 \times 10^6 = 0.38 \times 10^7\)

Add the terms in the numerator:
\(4.2 \times 10^7 + 0.38 \times 10^7 = 4.58 \times 10^7\)

Now divide by the denominator:
$$\frac{4.58 \times 10^7}{2.5 \times 10^{-3}} = \left(\frac{4.58}{2.5}\right) \times 10^{7 - (-3)}$$

Calculate the coefficient:
$$\frac{4.58}{2.5} = 1.832$$

Calculate the power of 10:
\(10^{7 + 3} = 10^{10}\)

Combine the results:
\(1.832 \times 10^{10}\)

M1: for adding the numerator to get \(4.58 \times 10^7\) (or \(45,800,000\)) or for showing a division of their numerator by \(2.5 \times 10^{-3}\)
M1: for a correct method to divide numbers in standard form, arriving at \(1.832 \times 10^k\) where \(k\) is an integer
A1: for \(1.832 \times 10^{10}\)

The probability of drawing two red counters is:
$$\text{P(Red, Red)} = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90}$$

The probability of drawing two blue counters is:
$$\text{P(Blue, Blue)} = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90}$$

The probability that both counters are of the same colour is:
$$\text{P(same colour)} = \text{P(Red, Red)} + \text{P(Blue, Blue)} = \frac{30}{90} + \frac{12}{90} = \frac{42}{90} = \frac{7}{15}$$

This can also be written as a decimal \(0.467\) (to 3 significant figures).

M1: for a correct probability for one of the outcomes, e.g., \(\frac{6}{10} \times \frac{5}{9}\) or \(\frac{4}{10} \times \frac{3}{9}\)
M1: for adding the two correct probabilities: \(\frac{30}{90} + \frac{12}{90}\)
A1: for \(\frac{7}{15}\) (or any equivalent fraction, or \(0.467\) or better)

First, find the ratio of the volumes of vase P and vase Q:
$$\frac{V_P}{V_Q} = \frac{108}{500} = \frac{27}{125}$$

Since the vases are mathematically similar, the ratio of their lengths is the cube root of the ratio of their volumes:
$$\frac{L_P}{L_Q} = \sqrt[3]{\frac{27}{125}} = \frac{3}{5}$$

The ratio of their surface areas is the square of the ratio of their lengths:
$$\frac{A_P}{A_Q} = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$$

We are given that the surface area of vase P is \(54 \text{ cm}^2\). We can set up the equation:
$$\frac{54}{A_Q} = \frac{9}{25}$$

Solve for \(A_Q\):
$$A_Q = 54 \times \frac{25}{9} = 6 \times 25 = 150 \text{ cm}^2$$

M1: for finding the linear scale factor \(\sqrt[3]{\frac{108}{500}} = \frac{3}{5}\) (or \(0.6\) or reciprocal)
M1: for finding the area scale factor \(\left(\frac{3}{5}\right)^2 = \frac{9}{25}\) (or \(0.36\) or reciprocal) and multiplying by 54 or dividing 54 by the factor
A1: for 150

To find the stationary points, we first find the derivative of \(y\) with respect to \(x\): \(\frac{dy}{dx} = 6x^2 - 18x - 24\). Setting \(\frac{dy}{dx} = 0\) for stationary points gives \(6x^2 - 18x - 24 = 0\). Dividing by 6, we get \(x^2 - 3x - 4 = 0\). Factoring the quadratic equation gives \((x - 4)(x + 1) = 0\), so the \(x\)-coordinates of the stationary points are \(x = 4\) and \(x = -1\). We substitute these values back into the original curve equation to find the corresponding \(y\)-coordinates: For \(x = 4\), \(y = 2(4)^3 - 9(4)^2 - 24(4) + 7 = -105\). For \(x = -1\), \(y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 7 = 20\). Therefore, the coordinates of the two stationary points are \((-1, 20)\) and \((4, -105)\).

M1: for differentiating at least one term correctly (e.g., \(6x^2\) or \(-18x\)). M1: for setting \(\frac{dy}{dx} = 0\) and finding the correct critical values \(x = 4\) and \(x = -1\). M1: for substituting at least one of their \(x\)-values back into the original equation to find a \(y\)-coordinate. A1.25: for both correct coordinates \((-1, 20)\) and \((4, -105)\).

First, multiply both sides of the equation by \(3w + 4\) to clear the fraction: \(t(3w + 4) = 5 - 2w\). Expand the left side: \(3wt + 4t = 5 - 2w\). Rearrange the terms to get all terms involving \(w\) on one side and all other terms on the other side: \(3wt + 2w = 5 - 4t\). Factor out \(w\) on the left side: \(w(3t + 2) = 5 - 4t\). Finally, divide by \(3t + 2\) to make \(w\) the subject: \(w = \frac{5 - 4t}{3t + 2}\).

M1: for multiplying both sides by \(3w + 4\) to get \(t(3w + 4) = 5 - 2w\). M1: for expanding the bracket and rearranging to gather all terms in \(w\) on one side: \(3wt + 2w = 5 - 4t\). M1: for factoring out \(w\): \(w(3t + 2) = 5 - 4t\). A1.25: for the correct final formula \(w = \frac{5 - 4t}{3t + 2}\).

The total number of counters in the bag is \(8 + 5 = 13\). The probability of taking at least one blue counter can be calculated as \(1 - P(\text{no blue counters})\), which is \(1 - P(\text{Red, Red})\). The probability of choosing a red counter first is \(\frac{8}{13}\). Since the counters are not replaced, the probability of choosing a second red counter is \(\frac{7}{12}\). Thus, \(P(\text{Red, Red}) = \frac{8}{13} \times \frac{7}{12} = \frac{56}{156} = \frac{14}{39}\). The probability of choosing at least one blue counter is \(1 - \frac{14}{39} = \frac{25}{39}\).

M1: for identifying that the required probability is \(1 - P(\text{Red, Red})\) or representing the three successful outcomes: \(P(\text{Red, Blue}) + P(\text{Blue, Red}) + P(\text{Blue, Blue})\). M1: for the product of two probabilities showing a correct non-replacement denominator: \(\frac{8}{13} \times \frac{7}{12}\). M1: for evaluating the calculation: \(1 - \frac{56}{156}\). A1.25: for the correct simplified fraction \(\frac{25}{39}\).

We can use the Cosine Rule to find the length of side \(AC\). The formula is: \(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle ABC)\). Substituting the given values: \(AC^2 = 7.5^2 + 11.2^2 - 2 \times 7.5 \times 11.2 \times \cos(68^\circ)\). Calculating the terms: \(AC^2 = 56.25 + 125.44 - 168 \times \cos(68^\circ)\). This simplifies to \(AC^2 = 181.69 - 168 \times 0.3746...\), so \(AC^2 \approx 181.69 - 62.934 = 118.756\). Taking the square root gives \(AC = \sqrt{118.756} \approx 10.897\text{ cm}\). Correct to 3 significant figures, the length of side \(AC\) is \(10.9\text{ cm}\).

M1: for correctly substituting into the Cosine Rule formula: \(7.5^2 + 11.2^2 - 2 \times 7.5 \times 11.2 \times \cos(68^\circ)\). M1: for evaluating parts of the expression, e.g., \(181.69 - 168 \times \cos(68^\circ)\). M1: for finding the value of \(AC^2\) as approximately \(118.7\) to \(119\) or demonstrating the square root step: \(\sqrt{118.756}\). A1.25: for the correct answer of \(10.9\) (accept answers in the range \(10.89\) to \(10.9\)).

The total volume is the sum of the volume of the cone and the volume of the hemisphere. The formula for the volume of a cone is \(V_{\text{cone}} = \frac{1}{3}\pi r^2 h\). Substituting \(r = 3\) and \(h = 9\): \(V_{\text{cone}} = \frac{1}{3}\pi (3)^2 (9) = 27\pi\text{ cm}^3\). The formula for the volume of a hemisphere is \(V_{\text{hemisphere}} = \frac{2}{3}\pi r^3\). Substituting \(r = 3\): \(V_{\text{hemisphere}} = \frac{2}{3}\pi (3)^3 = 18\pi\text{ cm}^3\). Adding these two volumes: \(V_{\text{total}} = 27\pi + 18\pi = 45\pi\text{ cm}^3\).

M1: for substituting correct values into the volume of a cone formula: \(\frac{1}{3}\pi \times 3^2 \times 9\). M1: for substituting correct values into the volume of a hemisphere formula: \(\frac{2}{3}\pi \times 3^3\). M1: for a method to add the two volumes in terms of \(\pi\): \(27\pi + 18\pi\). A1.25: for the final correct answer of \(45\pi\) (or exact decimal equivalent of \(141.37\) or \(141\)).

The area of a rectangle is length times width, so: \((2x + 5)(x - 1) = 47\). Expand the brackets: \(2x^2 - 2x + 5x - 5 = 47\), which simplifies to \(2x^2 + 3x - 5 = 47\). Subtract 47 from both sides to form a quadratic equation equal to zero: \(2x^2 + 3x - 52 = 0\). To solve this quadratic equation by factorisation, we find two numbers that multiply to \(2 \times (-52) = -104\) and add to \(3\). These numbers are \(13\) and \(-8\). Rewrite the middle term: \(2x^2 - 8x + 13x - 52 = 0\). Factor by grouping: \(2x(x - 4) + 13(x - 4) = 0\), which gives \((2x + 13)(x - 4) = 0\). This yields two solutions: \(x = 4\) or \(x = -6.5\). Since the width of the rectangle must be positive, \(x - 1 > 0\) implies \(x > 1\). Therefore, we reject \(x = -6.5\), and the correct value is \(x = 4\).

M1: for establishing the initial equation for the area: \((2x + 5)(x - 1) = 47\). M1: for expanding and rearranging to standard quadratic form: \(2x^2 + 3x - 52 = 0\). M1: for factorising to \((2x + 13)(x - 4) = 0\) or correctly applying the quadratic formula. A1.25: for \(x = 4\) as the only valid solution.

The ratio of the volumes of similar shapes is equal to the cube of the scale factor of their linear dimensions. Let \(k\) be the linear scale factor. We have: \(k^3 = \frac{\text{Volume of Jug } B}{\text{Volume of Jug } A} = \frac{1080}{320} = \frac{27}{8}\). To find the linear scale factor \(k\), we take the cube root of this ratio: \(k = \sqrt[3]{\frac{27}{8}} = \frac{3}{2} = 1.5\). Now, we multiply the height of jug \(A\) by this linear scale factor to get the height of jug \(B\): \(\text{Height of Jug } B = 12 \times 1.5 = 18\text{ cm}\).

M1: for finding the volume ratio: \(\frac{1080}{320}\) (or \(\frac{320}{1080}\) or simplified fraction \(\frac{27}{8}\)). M1: for taking the cube root of their volume ratio to find the linear scale factor: \(\sqrt[3]{\frac{27}{8}}\) or \(1.5\). M1: for multiplying the height of jug \(A\) by the linear scale factor: \(12 \times 1.5\). A1.25: for the correct height of \(18\).

To calculate the upper bound for the average speed, we use the formula: \(\text{Speed}_{\text{UB}} = \frac{d_{\text{UB}}}{t_{\text{LB}}}\). First, we find the upper bound of the distance \(d\). Since \(d = 400\) to the nearest 5 metres, \(d_{\text{UB}} = 400 + 2.5 = 402.5\text{ m}\). Next, we find the lower bound of the time \(t\). Since \(t = 48.4\) to the nearest 0.1 seconds, \(t_{\text{LB}} = 48.4 - 0.05 = 48.35\text{ s}\). Now, we compute the upper bound of the speed: \(\text{Speed}_{\text{UB}} = \frac{402.5}{48.35} \approx 8.324715\text{ m/s}\). Correct to 3 significant figures, this is \(8.32\text{ m/s}\).

M1: for finding the correct upper bound of \(d\) as \(402.5\) (or \(402.4\dot{9}\)). M1: for finding the correct lower bound of \(t\) as \(48.35\). M1: for dividing their upper bound of \(d\) by their lower bound of \(t\) (where \(400 < d_{\text{UB}} \le 402.5\) and \(48.35 \le t_{\text{LB}} < 48.4\)). A1.25: for the correct final answer of \(8.32\) (must be rounded to 3 significant figures).

**(a)** The volume of a cylinder is \(\pi r^2 h\) and the volume of a hemisphere is \(\frac{2}{3}\pi r^3\). Given that the total volume \(V = 300\pi\): \(\pi r^2 h + \frac{2}{3}\pi r^3 = 300\pi\). Dividing both sides by \(\pi\) gives: \(r^2 h + \frac{2}{3}r^3 = 300\). Rearranging to make \(h\) the subject: \(r^2 h = 300 - \frac{2}{3}r^3\) which gives \(h = \frac{300}{r^2} - \frac{2}{3}r\). The total surface area \(A\) of the solid consists of the curved surface area of the hemisphere, the curved surface area of the cylinder, and the flat circular base of the cylinder: \(A = 2\pi r^2 + 2\pi r h + \pi r^2 = 3\pi r^2 + 2\pi r h\). Substituting the expression for \(h\) into the surface area equation: \(A = 3\pi r^2 + 2\pi r\left(\frac{300}{r^2} - \frac{2}{3}r\right) = 3\pi r^2 + \frac{600\pi}{r} - \frac{4}{3}\pi r^2 = \frac{600\pi}{r} + \frac{5}{3}\pi r^2\). **(b)** To find the value of \(r\) for which \(A\) is a minimum, we differentiate \(A\) with respect to \(r\) and set the derivative to 0: \(A = 600\pi r^{-1} + \frac{5}{3}\pi r^2\), so \(\frac{dA}{dr} = -600\pi r^{-2} + \frac{10}{3}\pi r\). Setting \(\frac{dA}{dr} = 0\) gives: \(\frac{10}{3}\pi r = \frac{600\pi}{r^2}\). Dividing both sides by \(\pi\) and multiplying by \(r^2\): \(\frac{10}{3}r^3 = 600 \implies r^3 = 180\). Therefore, \(r = \sqrt[3]{180} \approx 5.6462...\). Correct to 3 significant figures, \(r = 5.65\) cm.

**Part (a)** M1: Formulates a correct volume equation \(\pi r^2 h + \frac{2}{3}\pi r^3 = 300\pi\) and attempts to make \(h\) the subject. M1: Writes down a correct expression for the total surface area \(A = 3\pi r^2 + 2\pi r h\). A1: Correctly substitutes \(h\) into \(A\) and simplifies with complete algebraic working shown to obtain the given expression. **Part (b)** M1: Differentiates the given expression for \(A\) with at least one term correct (either \(-600\pi r^{-2}\) or \(\frac{10}{3}\pi r\)). M1: Sets their derivative equal to 0 and rearranges to solve for \(r^3\) (obtaining \(r^3 = 180\)). A1: For 5.65 (accept 5.64 to 5.65).

First, we convert both numbers in the numerator to the same power of 10: \(4.2 \times 10^{7} + 0.39 \times 10^{7} = 4.59 \times 10^{7}\). Next, divide this by the denominator: \(\frac{4.59 \times 10^{7}}{1.5 \times 10^{-3}} = \frac{4.59}{1.5} \times 10^{7 - (-3)} = 3.06 \times 10^{10}\).

M1 for correctly converting the numerator to a single value, e.g., \(4.59 \times 10^{7}\) or \(45,900,000\). A1 for correct calculation of the division to obtain standard form: \(3.06 \times 10^{10}\).

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

M1 for factorising either the numerator as \((2x - 1)(x + 3)\) or the denominator as \((2x - 1)(2x + 1)\). A1 for final fully simplified fraction: \(\frac{x + 3}{2x + 1}\).

First, find \(g(1)\): \(g(1) = 2(1) + 5 = 7\). Next, find \(f(g(1)) = f(7)\): \(f(7) = \frac{4}{7 - 3} = \frac{4}{4} = 1\).

M1 for evaluating \(g(1) = 7\) or expressing \(fg(x) = \frac{4}{(2x + 5) - 3}\). A1 for final correct answer of 1.

To find the upper bound for \(A\), we use the upper bound of the numerator and the lower bound of the denominator: \(A_{\text{UB}} = \frac{b_{\text{UB}} - c_{\text{LB}}}{d_{\text{LB}}}\). Finding the bounds: \(b_{\text{UB}} = 12.65\), \(c_{\text{LB}} = 4.245\), \(d_{\text{LB}} = 2.35\). Substituting these values: \(A_{\text{UB}} = \frac{12.65 - 4.245}{2.35} = \frac{8.405}{2.35} \approx 3.57659\). Rounding to 3 significant figures gives \(3.58\).

M1 for identifying any of the correct bounds: \(b_{\text{UB}} = 12.65\), \(c_{\text{LB}} = 4.245\), or \(d_{\text{LB}} = 2.35\). M1 for the correct expression for upper bound of \(A\): \(\frac{12.65 - 4.245}{2.35}\). A1 for correct rounding to 3 significant figures: 3.58.

The sale price represents \(100\% - 15\% = 85\%\) of the normal price. Let \(x\) be the normal price: \(0.85x = 467.50\). Thus, \(x = \frac{467.50}{0.85} = 550\).

M1 for recognizing that \(85\%\) corresponds to \(467.50\) (e.g., writing \(0.85x = 467.50\) or \(467.50 \div 0.85\)). A1 for correct normal price of 550.

The total probability of choosing blue or yellow counters is \(1 - 0.35 = 0.65\). Since the ratio of blue to yellow is \(2:3\), the probability of choosing a blue counter is \(\frac{2}{2+3} = \frac{2}{5}\) of the remaining probability. Thus, \(P(\text{blue}) = \frac{2}{5} \times 0.65 = 0.26\).

M1 for calculating the combined probability of blue and yellow as \(1 - 0.35 = 0.65\). M1 for multiplying by the fraction representing blue: \(\frac{2}{5} \times 0.65\). A1 for the correct answer of 0.26.

First, calculate \(2\mathbf{a} - \mathbf{b} = 2\begin{pmatrix} 5 \\ -1 \end{pmatrix} - \begin{pmatrix} 4 \\ 6 \end{pmatrix} = \begin{pmatrix} 10 \\ -2 \end{pmatrix} - \begin{pmatrix} 4 \\ 6 \end{pmatrix} = \begin{pmatrix} 6 \\ -8 \end{pmatrix}\). Next, find the magnitude of this vector: \(|2\mathbf{a} - \mathbf{b}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\).

M1 for finding the vector \(2\mathbf{a} - \mathbf{b} = \begin{pmatrix} 6 \\ -8 \end{pmatrix}\). M1 for applying Pythagoras to find the magnitude: \(\sqrt{6^2 + (-8)^2}\). A1 for the correct answer of 10.

First, find the derivative of \(y\) with respect to \(x\): \(\frac{dy}{dx} = 6x - 12\). Substitute \(x = 5\) into the derivative: \(6(5) - 12 = 30 - 12 = 18\).

M1 for differentiating the equation of the curve to find \(\frac{dy}{dx} = 6x - 12\) (at least one term correct). M1 for substituting \(x = 5\) into their derivative. A1 for the correct gradient of 18.

Let \(P\) be the original price of the laptop. The price after a 15% increase is \(1.15P\). A 12% decrease is applied to this new price, which gives \(1.15P \times (1 - 0.12) = 1.15P \times 0.88\). We are given that the final price is £253, so \(1.15 \times 0.88 \times P = 253\). This simplifies to \(1.012P = 253\). Solving for \(P\), we get \(P = \frac{253}{1.012} = 250\).

M1 for writing an equation for the price changes, e.g., \(P \times 1.15 \times 0.88 = 253\) or finding the combined multiplier \(1.012\). A1 for correct original price of 250.

First, multiply both sides by \(5 - x\) to clear the fraction: \(y(5 - x) = 3x + 2\). Expand the left side: \(5y - xy = 3x + 2\). Rearrange the terms to group all terms with \(x\) on one side and terms without \(x\) on the other side: \(5y - 2 = 3x + xy\). Factor out \(x\) from the right side: \(5y - 2 = x(3 + y)\). Finally, divide both sides by \(3 + y\) to isolate \(x\): \(x = \frac{5y - 2}{3 + y}\).

M1 for multiplying by \(5 - x\) and expanding correctly: \(5y - xy = 3x + 2\). M1 for isolating terms in \(x\) on one side and factoring out \(x\): \(x(3 + y) = 5y - 2\). A1 for the correct final formula \(x = \frac{5y - 2}{y + 3} \) (or equivalent).

First, express both numbers in the numerator with the same power of 10: \(4.5 \times 10^7 + 12 \times 10^7 = 16.5 \times 10^7\), which is \(1.65 \times 10^8\). Next, divide this by the denominator: \(\frac{1.65 \times 10^8}{3 \times 10^{-4}} = \frac{1.65}{3} \times 10^{8 - (-4)} = 0.55 \times 10^{12}\). To write this in standard form, shift the decimal point to get \(5.5 \times 10^{11}\).

M1 for correctly simplifying the numerator to \(1.65 \times 10^8\) or \(165,000,000\). M1 for dividing by \(3 \times 10^{-4}\) to obtain \(5.5 \times 10^k\) where \(k\) is an integer. A1 for the correct answer \(5.5 \times 10^{11}\).

Using the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)\). Substitute the given values: \(AC^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \times \cos(60^\circ)\). Since \(\cos(60^\circ) = 0.5\), we have: \(AC^2 = 64 + 121 - 176 \times 0.5 = 185 - 88 = 97\). Thus, \(AC = \sqrt{97} \approx 9.8488578\text{ cm}\). Correct to 3 significant figures, the length of \(AC\) is \(9.85\text{ cm}\).

M1 for correct substitution into the Cosine Rule: \(8^2 + 11^2 - 2 \times 8 \times 11 \times \cos(60^\circ)\). M1 for evaluating \(AC^2 = 97\) or \(AC = \sqrt{97}\). A1 for correct value of 9.85 (accept 9.85 or 9.849).

The volume of a cylinder is given by \(V = \pi r^2 h\). Given \(V = 150\pi\) and \(h = 6\), we can set up the equation: \(150\pi = \pi r^2 \times 6\). Dividing both sides by \(6\pi\) gives \(r^2 = 25\), so the radius \(r = 5\text{ cm}\). The total surface area \(A\) of a cylinder is \(2\pi r^2 + 2\pi r h\). Substituting \(r = 5\) and \(h = 6\): \(A = 2\pi(5^2) + 2\pi(5)(6) = 2\pi(25) + 60\pi = 50\pi + 60\pi = 110\pi\text{ cm}^2\).

M1 for setting up the volume equation to find \(r\), e.g., \(150\pi = \pi r^2 \times 6\) leading to \(r = 5\). M1 for substituting \(r = 5\) and \(h = 6\) into the surface area formula \(2\pi r^2 + 2\pi r h\). A1 for \(110\pi\).

The total number of counters in the bag is \(5 + x + (x + 2) = 2x + 7\). The number of green counters is \(x + 2\). The probability of drawing a green counter is \(\frac{x + 2}{2x + 7}\). We are given that this probability is \(\frac{3}{7}\), so: \(\frac{x + 2}{2x + 7} = \frac{3}{7}\). Cross-multiplying gives: \(7(x + 2) = 3(2x + 7) \implies 7x + 14 = 6x + 21\). Solving for \(x\) gives \(x = 7\). Therefore, the total number of counters in the bag is \(2(7) + 7 = 21\).

M1 for writing an expression for the total counters, e.g., \(2x + 7\), and setting up the probability equation \(\frac{x + 2}{2x + 7} = \frac{3}{7}\). M1 for solving the equation to find \(x = 7\). A1 for 21.

Let the original number of boys be \(3x\) and the original number of girls be \(5x\). When 12 more girls join, the number of girls becomes \(5x + 12\). The new ratio is \(\frac{3x}{5x + 12} = \frac{1}{2}\). Cross-multiplying gives \(2(3x) = 1(5x + 12) \implies 6x = 5x + 12\). Solving for \(x\) gives \(x = 12\). The number of boys in the club is \(3x = 3 \times 12 = 36\).

M1 for representing the initial quantities as \(3x\) and \(5x\) and setting up the ratio equation \(\frac{3x}{5x + 12} = \frac{1}{2}\). M1 for solving to find \(x = 12\). A1 for the correct number of boys, 36.

To find the upper bound for \(A\), we need the upper bound for the numerator \(B\) and the lower bound for the denominator \(C\). The upper bound for \(B\) (45.7 correct to 1 d.p.) is \(B_{UB} = 45.75\). The lower bound for \(C\) (3.25 correct to 2 d.p.) is \(C_{LB} = 3.245\). Therefore, the upper bound for \(A\) is \(A_{UB} = \frac{45.75}{3.245} \approx 14.098613\). Rounding to 3 decimal places, we get \(14.099\).

M1 for identifying the correct bounds: \(B_{UB} = 45.75\) and \(C_{LB} = 3.245\). M1 for calculating \(\frac{45.75}{3.245}\). A1 for correct answer of 14.099 (must be to 3 d.p.).

To find the number of water drops needed, we divide the volume of the water tank by the volume of one drop: \text{Number of drops} = \frac{4.5 \times 10^4}{5 \times 10^{-5}}. We divide the coefficients and subtract the exponents: \frac{4.5}{5} = 0.9 and \( 10^4 \div 10^{-5} = 10^{4 - (-5)} = 10^9 \). So, the number of drops is: \( 0.9 \times 10^9 = 9 \times 10^8 \). This is in standard form.

M1: for showing a division of \( \frac{4.5 \times 10^4}{5 \times 10^{-5}} \) or an equivalent calculation. A1: for \( 9 \times 10^8 \) (accept other equivalent correct standard form notations like \( 9.0 \times 10^8 \)).

Let \( V \) be the original value of the machine. Each year, the machine retains \( 100\% - 8\% = 92\% \) of its value. So, after 3 years, the value of the machine is: \( V \times 0.92^3 = 19\,467.20 \). \( V \times 0.778688 = 19\,467.20 \). \( V = \frac{19\,467.20}{0.778688} = 25\,000 \). Therefore, the original value of the machine was \( £25\,000 \).

M1: for setting up the equation \( V \times 0.92^3 = 19\,467.20 \) or showing \( 19\,467.20 \div 0.92^3 \). A1: for \( 25\,000 \) (accept \( £25\,000 \)).

First, list the members of the universal set: \( \mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14\} \). Now, list the prime numbers less than 15: \( A = \{2, 3, 5, 7, 11, 13\} \). Now, list the odd numbers less than 15: \( B = \{1, 3, 5, 7, 9, 11, 13\} \). Find the union of \( A \) and \( B \): \( A \cup B = \{1, 2, 3, 5, 7, 9, 11, 13\} \). The complement set \( (A \cup B)' \) contains all members of \( \mathcal{E} \) that are not in \( A \cup B \): \( (A \cup B)' = \{4, 6, 8, 10, 12, 14\} \).

M1: for listing at least 4 elements of \( A \cup B \) or listing \( (A \cup B)' \) with at most 1 missing or 1 extra number. A1: for the correct list: \( 4, 6, 8, 10, 12, 14 \) (order does not matter, brackets are optional).

To find the gradient of the curve, we differentiate the equation \( y \) with respect to \( x \): \( \frac{dy}{dx} = 6x^2 - 10x \). Now substitute \( x = 2 \) into the derivative: \( \text{Gradient} = 6(2)^2 - 10(2) = 24 - 20 = 4 \).

M1: for differentiating at least one term correctly (e.g., \( 6x^2 \) or \( -10x \)). A1: for the correct derivative \( 6x^2 - 10x \). A1: for \( 4 \) from correct working.

1. **Find the total surface area of the cylinder:**
The total surface area of a cylinder is given by:
\(A_c = 2\pi r^2 + 2\pi r h\)
Given that \(h = 2r\):
\(A_c = 2\pi r^2 + 2\pi r (2r) = 2\pi r^2 + 4\pi r^2 = 6\pi r^2\)

2. **Find the surface area of the sphere:**
The surface area of a sphere is given by:
\(A_s = 4\pi R^2\)

3. **Equate the surface areas to find a relation between \(r\) and \(R\):**
\(6\pi r^2 = 4\pi R^2\)
\(R^2 = \frac{6\pi}{4\pi} r^2 = \frac{3}{2} r^2\)
\(R = \sqrt{\frac{3}{2}} r\)

4. **Find the volume of the cylinder (\(V_c\)):**
\(V_c = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3\)

5. **Find the volume of the sphere (\(V_s\)):**
\(V_s = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi \left(\sqrt{\frac{3}{2}}r\right)^3 = \frac{4}{3}\pi \left(\frac{3}{2}\sqrt{\frac{3}{2}}\right) r^3 = 2\sqrt{\frac{3}{2}}\pi r^3\)

6. **Find the ratio of \(V_c : V_s\):**
\(V_c : V_s = 2\pi r^3 : 2\sqrt{\frac{3}{2}}\pi r^3\)
\(V_c : V_s = 1 : \sqrt{\frac{3}{2}}\)
Multiply both parts of the ratio by \(2\):
\(2 : 2\sqrt{\frac{3}{2}} = 2 : \sqrt{6}\)
To write this in the form \(\sqrt{a} : b\), multiply both parts by \(\sqrt{6}\):
\(2\sqrt{6} : 6\)
Divide both parts by \(2\):
\(\sqrt{6} : 3\)

Here, \(a = 6\) and \(b = 3\).

M1: Method to find total surface area of the cylinder in terms of \(r\) (e.g., \(6\pi r^2\)).
M1: Equating \(6\pi r^2 = 4\pi R^2\) and solving to find a relation between \(r\) and \(R\) (e.g., \(R = \sqrt{1.5}r\)).
M1: Correct expression for the volume of the cylinder (\(2\pi r^3\)) and volume of the sphere in terms of \(r\) (e.g., \(2\sqrt{1.5}\pi r^3\)).
M1: Forming the ratio \(V_c : V_s\) and attempting to simplify.
A1: Correct final ratio in the required form: \(\sqrt{6} : 3\) (or equivalent values of \(a = 6, b = 3\)).

1. **Factorise the first numerator:**
\(2x^2 - 5x - 12 = 2x^2 - 8x + 3x - 12 = 2x(x - 4) + 3(x - 4) = (2x + 3)(x - 4)\)

2. **Factorise the first denominator:**
\(3x^2 - 12x = 3x(x - 4)\)

3. **Factorise the second numerator:**
\(2x^2 + 7x + 6 = 2x^2 + 4x + 3x + 6 = 2x(x + 2) + 3(x + 2) = (2x + 3)(x + 2)\)

4. **Factorise the second denominator:**
\(x^2 - 4 = (x - 2)(x + 2)\)

5. **Rewrite the division as multiplication by the reciprocal:**
\[\frac{(2x + 3)(x - 4)}{3x(x - 4)} \times \frac{(x - 2)(x + 2)}{(2x + 3)(x + 2)}\]

6. **Cancel common terms:**
- Cancel \((x - 4)\) from the first fraction:
\[\frac{2x + 3}{3x} \times \frac{(x - 2)(x + 2)}{(2x + 3)(x + 2)}\]
- Cancel \((x + 2)\) from the second fraction:
\[\frac{2x + 3}{3x} \times \frac{x - 2}{2x + 3}\]
- Cancel \((2x + 3)\) from numerator and denominator:
\[\frac{x - 2}{3x}\]

M1: Factorising the first numerator to \((2x + 3)(x - 4)\) or the second numerator to \((2x + 3)(x + 2)\).
M1: Factorising the first denominator to \(3x(x - 4)\) and the second denominator to \((x - 2)(x + 2)\).
M1: Complete method of multiplying by the reciprocal and attempting to cancel common factors.
A1: \(\frac{x - 2}{3x}\) (or equivalent single fraction, e.g., \(\frac{1}{3} - \frac{2}{3x}\)).

**Part (a):**
1. The number of red counters is \(x\).
The number of blue counters is \(10 - x\).

2. The probability of choosing two red counters (without replacement) is:
\[P(\text{Red, Red}) = \frac{x}{10} \times \frac{x - 1}{9} = \frac{x(x - 1)}{90}\]

3. The probability of choosing two blue counters (without replacement) is:
\[P(\text{Blue, Blue}) = \frac{10 - x}{10} \times \frac{9 - x}{9} = \frac{(10 - x)(9 - x)}{90}\]

4. The probability that both counters are the same colour is:
\[P(\text{Same}) = \frac{x(x - 1) + (10 - x)(9 - x)}{90} = \frac{7}{15}\]

5. Expand and simplify the numerator:
\[x^2 - x + 90 - 19x + x^2 = 2x^2 - 20x + 90\]
So,
\[\frac{2x^2 - 20x + 90}{90} = \frac{7}{15}\]

6. Multiply both sides by 90:
\[2x^2 - 20x + 90 = 42\]
\[2x^2 - 20x + 48 = 0\]

7. Divide the entire equation by 2:
\[x^2 - 10x + 24 = 0\] (Shown)

**Part (b):**
1. Solve the quadratic equation to find possible values of \(x\):
\[(x - 6)(x - 4) = 0\implies x = 6 \text{ or } x = 4\]

2. Since there are more red counters than blue counters, \(x > 5\). Therefore, \(x = 6\).
So there are 6 red counters and 4 blue counters.

3. The probability that at least one of the counters is blue is:
\[1 - P(\text{both red}) = 1 - \frac{6}{10} \times \frac{5}{9} = 1 - \frac{30}{90} = 1 - \frac{1}{3} = \frac{2}{3}\]

M1: Writing correct probability terms for \(P(\text{Red, Red})\) or \(P(\text{Blue, Blue})\) in terms of \(x\).
M1: Equating the sum of probabilities to \(\frac{7}{15}\) and attempting to expand.
A1: (Part a) Complete algebraic progression leading to \(x^2 - 10x + 24 = 0\) with no errors.
M1: (Part b) Solving the quadratic to find \(x = 6\) and identifying that there are 6 red and 4 blue counters.
A1: (Part b) Correct probability of \(\frac{2}{3}\) (or equivalent fraction / decimal \(0.667\)).

1. **Find the first derivative of \(y\):**
\[\frac{dy}{dx} = 6x^2 - 18x - 24\]

2. **Set \(\frac{dy}{dx} = 0\) to find the \(x\)-coordinates of stationary points:**
\[6x^2 - 18x - 24 = 0\]
Divide by 6:
\[x^2 - 3x - 4 = 0\]
Factorise:
\[(x - 4)(x + 1) = 0\implies x = 4\text{ or } x = -1\]

3. **Find the corresponding \(y\)-coordinates:**
- For \(x = -1\):
\[y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 15 = -2 - 9 + 24 + 15 = 28\]
So, one stationary point is \((-1, 28)\).

- For \(x = 4\):
\[y = 2(4)^3 - 9(4)^2 - 24(4) + 15 = 128 - 144 - 96 + 15 = -97\]
So, the other stationary point is \((4, -97)\).

4. **Determine the nature of each point using the second derivative:**
\[\frac{d^2y}{dx^2} = 12x - 18\]

- At \(x = -1\):
\[\frac{d^2y}{dx^2} = 12(-1) - 18 = -30\]
Since \(-30 < 0\), the point \((-1, 28)\) is a **local maximum**.

- At \(x = 4\):
\[\frac{d^2y}{dx^2} = 12(4) - 18 = 48 - 18 = 30\]
Since \(30 > 0\), the point \((4, -97)\) is a **local minimum**.

M1: Correct differentiation to obtain \(\frac{dy}{dx} = 6x^2 - 18x - 24\).
M1: Setting \(\frac{dy}{dx} = 0\) and finding the correct \(x\)-values (\(x = 4\) and \(x = -1\)).
M1: Substituting both \(x\)-values into the original equation to obtain \(y = 28\) and \(y = -97\).
M1: Finding \(\frac{d^2y}{dx^2} = 12x - 18\) and substituting at least one of the \(x\)-values.
A1: Identifying \((-1, 28)\) as a maximum and \((4, -97)\) as a minimum with fully correct supporting work.

**Part (a):**
1. Let \(y = \frac{3}{x - 2}\).
2. Rearrange the formula to make \(x\) the subject:
\[y(x - 2) = 3\]
\[xy - 2y = 3\]
\[xy = 2y + 3\]
\[x = \frac{2y + 3}{y}\]
3. Replace \(y\) with \(x\) to obtain the inverse function:
\[f^{-1}(x) = \frac{2x + 3}{x}\]

**Part (b):**
1. Find the composite function \(fg(x)\):
\[fg(x) = f(g(x)) = f(2x + 1) = \frac{3}{(2x + 1) - 2} = \frac{3}{2x - 1}\]
2. Set \(fg(x) = \frac{1}{2}\):
\[\frac{3}{2x - 1} = \frac{1}{2}\]
3. Cross-multiply and solve for \(x\):
\[6 = 2x - 1\]
\[2x = 7\]
\[x = 3.5\]

M1: (Part a) Attempting to rearrange \(y = \frac{3}{x - 2}\) to make \(x\) the subject.
A1: (Part a) Correct expression \(f^{-1}(x) = \frac{2x + 3}{x}\) (or equivalent form, e.g., \(2 + \frac{3}{x}\)).
M1: (Part b) Correctly substituting \(g(x)\) into \(f(x)\) to get \(\frac{3}{2x - 1} = \frac{1}{2}\) (or solving \(g(x) = f^{-1}(1/2)\)).
A1: (Part b) Correct value of \(x = 3.5\) (or \(\frac{7}{2}\)).

1. **Identify the projection of the line \(EC\) onto the base \(ABCD\):**
Since \(E\) is vertically above \(A\), the projection of the point \(E\) onto the base \(ABCD\) is \(A\).
Therefore, the projection of the line \(EC\) onto the horizontal base is the line \(AC\).
The angle we need to calculate is \(\angle ECA\).

2. **Calculate the length of the diagonal \(AC\) on the base:**
\(ABCD\) is a rectangle with \(AB = 12\text{ cm}\) and \(BC = AD = 8\text{ cm}\).
Using Pythagoras' theorem in the right-angled triangle \(ABC\):
\[AC^2 = AB^2 + BC^2\]
\[AC^2 = 12^2 + 8^2 = 144 + 64 = 208\]
\[AC = \sqrt{208}\text{ cm} \approx 14.4222\text{ cm}\]

3. **Use trigonometry in the right-angled triangle \(EAC\):**
Since \(EA\) is vertical and \(AC\) lies in the horizontal base, \(\angle EAC = 90^\circ\).
In triangle \(EAC\):
\[\tan(\angle ECA) = \frac{EA}{AC}\]
\[\tan(\angle ECA) = \frac{5}{\sqrt{208}}\]
\[\angle ECA = \tan^{-1}\left(\frac{5}{\sqrt{208}}\right)\]
\[\angle ECA \approx \tan^{-1}(0.346688) \approx 19.1206^\circ\]

4. **Round to 1 decimal place:**
The angle is \(19.1^\circ\).

M1: Correct use of Pythagoras' theorem to find \(AC^2\) or \(AC\) (e.g., \(12^2 + 8^2\)).
A1: Finding \(AC = \sqrt{208}\) or \(14.4\) (or better).
M1: Identifying the required angle as \(\angle ECA\) (or visualising the correct right-angled triangle \(EAC\)).
M1: Setting up a correct trigonometric ratio for \(\angle ECA\) (e.g., \(\tan(\theta) = \frac{5}{\sqrt{208}}\)).
A1: Correct final answer of \(19.1\) (accept range \(19.1 - 19.2\)).

Let \(V\) be the volume of the box:
\[V = x^2 h = 108\]

Express \(h\) in terms of \(x\):
\[h = \frac{108}{x^2}\]

The total surface area \(A\) of the open-topped box consists of the square base and four vertical sides:
\[A = x^2 + 4xh\]

Substitute the expression for \(h\) into the surface area formula:
\[A = x^2 + 4x\left(\frac{108}{x^2}\right) = x^2 + \frac{432}{x}\]

To find the minimum surface area, differentiate \(A\) with respect to \(x\):
\[\frac{dA}{dx} = 2x - 432x^{-2} = 2x - \frac{432}{x^2}\]

Set the derivative to zero to find the stationary point:
\[2x - \frac{432}{x^2} = 0\]
\[2x = \frac{432}{x^2}\]
\[2x^3 = 432\]
\[x^3 = 216\]
\[x = 6\]

To confirm this gives a minimum, find the second derivative:
\[\frac{d^2A}{dx^2} = 2 + 864x^{-3} = 2 + \frac{864}{x^3}\]

Since \(x = 6\):
\[\frac{d^2A}{dx^2} = 2 + \frac{864}{216} = 2 + 4 = 6 > 0\]

Since the second derivative is positive, \(x = 6\) gives a minimum.

Now calculate the minimum surface area:
\[A = 6^2 + \frac{432}{6} = 36 + 72 = 108\text{ cm}^2\]

**M1**: Writes an expression for the volume of the box, e.g., \(x^2 h = 108\), or expresses \(h\) in terms of \(x\) as \(h = \frac{108}{x^2}\).

**M1**: Substitutes the expression for \(h\) into the total surface area formula, \(A = x^2 + 4xh\), to get \(A = x^2 + \frac{432}{x}\).

**M1**: Correctly differentiates at least one term of \(A\) with respect to \(x\), establishing \(\frac{dA}{dx} = 2x - \frac{432}{x^2}\).

**M1**: Sets the derivative to 0 and solves for \(x\) to obtain \(x = 6\) (or shows \(x^3 = 216\)).

**M1**: Verifies that \(x = 6\) yields a minimum, e.g., by evaluating \(\frac{d^2A}{dx^2} = 2 + \frac{864}{x^3}\) at \(x = 6\) to get \(6 > 0\).

**A1**: For the correct minimum surface area of \(108\).

The total number of sweets in the bag is \(n\).

The number of lemon sweets is \(6\).

The number of strawberry sweets is \(n - 6\).

The probability of selecting two lemon sweets is:
\[P(\text{Lemon, Lemon}) = \frac{6}{n} \times \frac{5}{n-1} = \frac{30}{n(n-1)}\]

The probability of selecting two strawberry sweets is:
\[P(\text{Strawberry, Strawberry}) = \frac{n-6}{n} \times \frac{n-7}{n-1} = \frac{(n-6)(n-7)}{n(n-1)}\]

The probability that both sweets are of the same flavor is the sum of these two probabilities:
\[P(\text{Same flavor}) = \frac{30 + (n-6)(n-7)}{n(n-1)} = \frac{1}{2}\]

Expand the numerator:
\[(n-6)(n-7) = n^2 - 13n + 42\]

Substitute this back into the equation:
\[\frac{30 + n^2 - 13n + 42}{n(n-1)} = \frac{1}{2}\]
\[\frac{n^2 - 13n + 72}{n^2 - n} = \frac{1}{2}\]

Multiply both sides by \(2(n^2 - n)\):
\[2(n^2 - 13n + 72) = n^2 - n\]
\[2n^2 - 26n + 144 = n^2 - n\]

Rearrange to form a quadratic equation equal to zero:
\[n^2 - 25n + 144 = 0\]

Solve the quadratic equation by factoring:
\[(n-16)(n-9) = 0\]

This gives two possible solutions:
\[n = 16 \quad \text{or} \quad n = 9\]

Since the question states that \(n > 10\), we reject \(n = 9\).

Therefore, \(n = 16\).

**M1**: Writes a correct expression for the probability of selecting two lemon sweets, \(\frac{30}{n(n-1)}\), or two strawberry sweets, \(\frac{(n-6)(n-7)}{n(n-1)}\).

**M1**: Sets up a correct equation for the sum of the probabilities equaling \(\frac{1}{2}\), i.e., \(\frac{30 + (n-6)(n-7)}{n(n-1)} = \frac{1}{2}\).

**M1**: Expands the numerator and denominator correctly to get \(\frac{n^2 - 13n + 72}{n^2 - n} = \frac{1}{2}\).

**M1**: Rearranges the equation to a quadratic form, e.g., \(2n^2 - 26n + 144 = n^2 - n\).

**M1**: Correctly solves the simplified quadratic equation \(n^2 - 25n + 144 = 0\) to find \(n = 16\) and \(n = 9\).

**A1**: For concluding that \(n = 16\) because \(n > 10\).