2023 Edexcel IGCSE Mathematics (Specification A) 模擬試題連答案詳解

We expand the brackets by multiplying each term in the first bracket by each term in the second: \((3x)(2x) + (3x)(5) + (-4)(2x) + (-4)(5) = 6x^2 + 15x - 8x - 20\). Simplifying the middle terms gives \(6x^2 + 7x - 20\).

M1 for expanding at least two terms correctly (e.g. \(6x^2\) and \(15x\)). A0.5 for the fully simplified correct expression \(6x^2 + 7x - 20\).

The sum of the probabilities of all mutually exclusive events is 1. Therefore, the probability of choosing a yellow counter is \(1 - (0.35 + 0.4) = 1 - 0.75 = 0.25\).

M1 for \(1 - (0.35 + 0.4)\) or showing the sum of red and blue probabilities is \(0.75\). A0.5 for \(0.25\) (or equivalent fraction).

A \(10\%\) discount means the laptop is sold for \(90\%\) of its normal price. Let the normal price be \(x\). We have \(0.90x = 540\), which gives \(x = \frac{540}{0.90} = 600\).

M1 for setting up \(0.90x = 540\) or doing \(540 \div 0.9\). A0.5 for \(600\).

By Pythagoras' theorem, \(AC^2 = AB^2 + BC^2\). Substituting the given values: \(AC^2 = 6^2 + 8^2 = 36 + 64 = 100\). Taking the square root: \(AC = \sqrt{100} = 10\text{ cm}\).

M1 for applying Pythagoras' theorem, e.g. \(6^2 + 8^2\) or \(\sqrt{100}\). A0.5 for \(10\).

First, find the total number of parts in the ratio: \(2 + 5 = 7\) parts. Next, find the value of one part: \(350 \div 7 = 50\). Finally, find the larger share (which represents 5 parts): \(5 \times 50 = 250\).

M1 for dividing 350 by 7, or for finding one part is 50, or writing \(\frac{5}{7} \times 350\). A0.5 for \(250\).

To write \(0.00045\) in standard form, move the decimal point 4 places to the right to get a number between 1 and 10, which is \(4.5\). Since the decimal point moved to the right, the power of 10 is negative: \(4.5 \times 10^{-4}\).

M1 for \(4.5 \times 10^{n}\) where \(n\) is a non-zero integer. A0.5 for \(4.5 \times 10^{-4}\).

The radius of the circle is half of the diameter, so \(r = 12 \div 2 = 6\text{ cm}\). The formula for the area of a circle is \(A = \pi r^2\). Substituting the radius gives \(A = \pi \times 6^2 = 36\pi\text{ cm}^2\).

M1 for finding radius is \(6\) and substituting into area formula, e.g. \(\pi \times 6^2\). A0.5 for \(36\pi\).

Add \(7\) to both sides of the inequality: \(4x < 20\). Divide both sides by \(4\) to find the solution: \(x < 5\).

M1 for isolating the term with \(x\), e.g. \(4x < 20\). A0.5 for \(x < 5\).

To factorise \(12x^2y - 18xy^2\) fully: 1. Find the highest common factor (HCF) of the numerical coefficients 12 and 18, which is 6. 2. Find the highest common factor of the algebraic terms: for \(x^2\) and \(x\), the HCF is \(x\); for \(y\) and \(y^2\), the HCF is \(y\). 3. Combining these gives the overall common factor as \(6xy\). 4. Divide each term of the original expression by \(6xy\) to find the terms inside the bracket: \(\frac{12x^2y}{6xy} = 2x\) and \(\frac{18xy^2}{6xy} = 3y\). 5. Thus, the fully factorised expression is \(6xy(2x - 3y)\).

M1 for finding a partial common factor (e.g. \(2xy(6x - 9y)\), \(6x(2xy - 3y^2)\), or \(6xy(...)\) with one incorrect term inside). A0.5 for the fully correct factorised expression \(6xy(2x - 3y)\).

We are given the ratios: \(B : R = 5 : 3\) and \(R : G = 4 : 7\). To find the ratio of blue pens to green pens, we make the parts for red pens equal in both ratios. The lowest common multiple of 3 and 4 is 12. Multiply the first ratio by 4: \(B : R = 20 : 12\). Multiply the second ratio by 3: \(R : G = 12 : 21\). Now we can write the combined ratio as \(B : R : G = 20 : 12 : 21\). Therefore, the ratio of blue pens to green pens is \(B : G = 20 : 21\). Since 20 and 21 have no common factors other than 1, this ratio is in its simplest form.

M1 for a valid method to scale the ratios to have a common value for red pens (e.g. writing \(20:12\) and \(12:21\), or calculating \(\frac{5}{3} \times \frac{4}{7}\)). A0.5 for the correct final ratio of \(20:21\) (or equivalent).

1. First, calculate the selling price before the sale reduction (with a \(15\%\) profit): \(\text{Selling Price} = \$80 \times (1 + 0.15) = \$80 \times 1.15 = \$92\). 2. Next, calculate the final sale price after a \(10\%\) reduction on the selling price: \(\text{Sale Price} = \$92 \times (1 - 0.10) = \$92 \times 0.9 = \$82.80\).

M1 for finding the intermediate selling price of \(\$92\) or for setting up a complete correct multiplication expression such as \(80 \times 1.15 \times 0.9\). A0.5 for the correct final price of \(\$82.80\) (or \(82.8\)).

Method 1: Using prime factors. Find the prime factorisation of both numbers: \(56 = 2^3 \times 7\) and \(84 = 2^2 \times 3 \times 7\). To find the LCM, take the highest power of each prime factor that appears in either number: for 2, the highest power is \(2^3\); for 3, the highest power is \(3^1\); for 7, the highest power is \(7^1\). Thus, \(\text{LCM} = 2^3 \times 3 \times 7 = 8 \times 3 \times 7 = 168\). Method 2: Listing multiples. Multiples of 56 are: 56, 112, 168, 224, ... Multiples of 84 are: 84, 168, 252, ... The first common multiple is 168.

M1 for listing at least three multiples of both numbers (containing 168 in both lists) or for showing the prime factorisation of both numbers. A0.5 for the correct final answer of 168.

First, expand the bracket on the left-hand side:
\(4x - 12 \ge 7x - 18\)

Next, rearrange to collect the terms in \(x\) on one side and the constant terms on the other. Subtract \(4x\) from both sides:
\(-12 \ge 3x - 18\)

Add 18 to both sides:
\(6 \ge 3x\)

Divide both sides by 3:
\(2 \ge x\)

Which can be written as:
\(x \le 2\)

M1: for expanding the bracket correctly to get \(4x - 12\)
M1: for a correct process to isolate \(x\) on one side (e.g. \(3x \le 6\) or \(-3x \ge -6\))
A1: for \(x \le 2\) (or \(2 \ge x\))

Since the probability of picking a green counter is \(0.2\), the probability of picking either a red counter or a blue counter is:
\(1 - 0.2 = 0.8\)

The ratio of the number of red counters to blue counters is \(3 : 5\).
This means the probability is shared in the ratio \(3 : 5\), which has a total of \(3 + 5 = 8\) parts.

The probability of picking a blue counter is:
\(\frac{5}{8} \times 0.8 = 5 \times 0.1 = 0.5\)

M1: for finding the combined probability of red and blue as \(1 - 0.2 = 0.8\)
M1: for sharing the combined probability in the ratio \(3 : 5\), e.g., \(\frac{5}{8} \times 0.8\)
A1: for \(0.5\) (or equivalent fraction such as \(\frac{1}{2\}})

In the right-angled triangle \(ABC\), relative to the angle \(BAC\):
- \(AB\) is the adjacent side (\(7.2\text{ cm}\))
- \(AC\) is the hypotenuse (\(13.5\text{ cm}\))

Using the cosine ratio:
\(\cos(BAC) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{7.2}{13.5}\)

Find the angle by taking the inverse cosine:
\(BAC = \arccos\left(\frac{7.2}{13.5}\right)\)
\(BAC \approx 57.769^\circ\)

Rounding to 1 decimal place gives:
\(57.8^\circ\)

M1: for recognizing the correct trigonometric ratio, e.g., \(\cos(BAC) = \frac{7.2}{13.5}\) or \(\cos(x) = 0.533...\)
M1: for a correct process to find the angle, e.g., \(\arccos\left(\frac{7.2}{13.5}\right)\)
A1: for \(57.8\) (accept \(57.8^\circ\))

A reduction of \(15\%\) means the sale price represents \(100\% - 15\% = 85\%\) of the normal price.

Let \(x\) be the normal price:
\(0.85 \times x = 74.80\)

Solve for \(x\):
\(x = \frac{74.80}{0.85} = 88\)

The normal price of the jacket is \(£88\).

M1: for establishing that \(85\%\) corresponds to \(74.80\), e.g., \(0.85 \times \text{normal price} = 74.80\) or \(\frac{74.80}{85}\)
M1: for a complete method to find the normal price, e.g., \(\frac{74.80}{0.85}\) or \(\frac{74.80}{85} \times 100\)
A1: for \(88\) (or \(£88\))

First, find the midpoint (\(x\)) for each interval:
- For \(0 < t \le 5\), midpoint = \(2.5\)
- For \(5 < t \le 10\), midpoint = \(7.5\)
- For \(10 < t \le 15\), midpoint = \(12.5\)
- For \(15 < t \le 20\), midpoint = \(17.5\)
- For \(20 < t \le 25\), midpoint = \(22.5\)

Next, multiply each midpoint by its corresponding frequency (\(f \times x\)):
- \(2.5 \times 8 = 20\)
- \(7.5 \times 13 = 97.5\)
- \(12.5 \times 11 = 137.5\)
- \(17.5 \times 5 = 87.5\)
- \(22.5 \times 3 = 67.5\)

Sum of frequencies (\(\sum f\)) = \(40\)
Sum of products (\(\sum fx\)) = \(20 + 97.5 + 137.5 + 87.5 + 67.5 = 410\)

Estimate of the mean = \(\frac{\sum fx}{\sum f} = \frac{410}{40} = 10.25\) minutes.

M1: for finding the midpoints of the intervals (at least 4 correct midpoints shown or used)
M1: for a correct method to find the sum of \(f \times x\) (using midpoints) and dividing by 40, e.g., \(\frac{\sum fx}{40}\)
A1: for \(10.25\) (or equivalent fraction like \(\frac{41}{4}\))

Initially, the ratio of part-time to full-time is \(3 : 7\).
We are given there are 112 full-time workers.
Using this ratio, we find the number of part-time workers:
\(\text{Part-time} = 112 \times \frac{3}{7} = 16 \times 3 = 48\)

Now, the new ratio of part-time : full-time : temporary is \(6 : 14 : 5\).
The number of part-time (48) and full-time (112) workers remains unchanged.
Let's find the multiplier for the new ratio:
\(\frac{48}{6} = 8\) (or \(\frac{112}{14} = 8\))

So, the number of temporary workers hired is:
\(5 \times 8 = 40\)

M1: for finding the number of part-time workers, e.g., \(112 \times \frac{3}{7} = 48\)
M1: for a correct process to find the multiplier for the new ratio or relating the ratio parts, e.g., \(\frac{48}{6} = 8\) or \(\frac{112}{14} = 8\)
A1: for \(40\)

To solve the equation, we first multiply all terms by the common denominator, which is \(x(x-2)\):

$$8x - 6(x-2) = 1 \cdot x(x-2)$$

Expand both sides:

$$8x - 6x + 12 = x^2 - 2x$$

$$2x + 12 = x^2 - 2x$$

Rearrange the terms into a standard quadratic equation format:

$$x^2 - 4x - 12 = 0$$

Factorise the quadratic equation:

$$(x-6)(x+2) = 0$$

This gives the solutions:

$$x = 6 \quad \text{or} \quad x = -2$$

M1: For removing fractions by multiplying by \(x(x-2)\), e.g., \(8x - 6(x-2) = x(x-2)\)
M1: For expanding and simplifying to a 3-term quadratic equation, e.g., \(x^2 - 4x - 12 = 0\)
M1: For a correct method to solve their 3-term quadratic equation (e.g., factorisation to \((x-6)(x+2) = 0\))
A1: For both correct solutions: \(x = 6\) and \(x = -2\)

First, find the derivative of the curve equation with respect to \(x\):

$$\frac{dy}{dx} = 6x^2 - 18x - 24$$

At stationary points, the gradient is zero, so we set \(\frac{dy}{dx} = 0\):

$$6x^2 - 18x - 24 = 0$$

Divide the entire equation by 6:

$$x^2 - 3x - 4 = 0$$

Factorise the quadratic equation:

$$(x-4)(x+1) = 0$$

Thus, the \(x\)-coordinates are \(x = 4\) and \(x = -1\).

Now, substitute these \(x\)-values back into the original curve equation to find the corresponding \(y\)-coordinates:

For \(x = 4\):

$$y = 2(4)^3 - 9(4)^2 - 24(4) + 15 = 128 - 144 - 96 + 15 = -97$$

For \(x = -1\):

$$y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 15 = -2 - 9 + 24 + 15 = 28$$

The coordinates of the stationary points are \((4, -97)\) and \((-1, 28)\).

M1: For an attempt to differentiate, with at least two terms correct (e.g., \(6x^2 - 18x\))
M1: For setting their derivative to 0 and attempting to solve the resulting quadratic equation
A1: For finding correct \(x\)-values of \(x = 4\) and \(x = -1\)
A1: For correct coordinates \((4, -97)\) and \((-1, 28)\)

First, calculate the angle \(BAC\) between the lines \(AB\) and \(AC\):

$$\angle BAC = 135^\circ - 60^\circ = 75^\circ$$

We can now use the Cosine Rule in triangle \(ABC\) to find the distance \(BC\):

$$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)$$

Substitute the known values into the formula:

$$BC^2 = 12^2 + 15^2 - 2(12)(15)\cos(75^\circ)$$

$$BC^2 = 144 + 225 - 360\cos(75^\circ)$$

$$BC^2 = 369 - 360(0.258819...)$$

$$BC^2 = 369 - 93.1748...$$

$$BC^2 = 275.825...$$

$$BC = \sqrt{275.825...} \approx 16.608\text{ km}$$

Correct to 3 significant figures, the distance is \(16.6\text{ km}\).

M1: For calculating the included angle: \(\angle BAC = 135^\circ - 60^\circ = 75^\circ\)
M1: For substituting correctly into the Cosine Rule: \(BC^2 = 12^2 + 15^2 - 2 \times 12 \times 15 \times \cos(75^\circ)\)
A1: For evaluating \(BC^2\) correctly as approximately \(275.8\) or \(BC \approx 16.61\)
A1: For the final answer of \(16.6\) (accept answers in range 16.6 to 16.61)

The total number of pens in the box initially is \(7 + 5 = 12\).

There are two scenarios where the two pens are of different colours:
1. First pen is blue, second pen is red: \(P(\text{Blue, Red})\)
2. First pen is red, second pen is blue: \(P(\text{Red, Blue})\)

Calculate the probability for each scenario:

$$P(\text{Blue, Red}) = \frac{7}{12} \times \frac{5}{11} = \frac{35}{132}$$

$$P(\text{Red, Blue}) = \frac{5}{12} \times \frac{7}{11} = \frac{35}{132}$$

To find the overall probability of getting different colours, add these two probabilities together:

$$P(\text{Different Colours}) = \frac{35}{132} + \frac{35}{132} = \frac{70}{132}$$

Simplify the fraction by dividing the numerator and denominator by 2:

$$\frac{70}{132} = \frac{35}{66}$$

M1: For identifying both combinations: Blue-Red and Red-Blue
M1: For a correct probability calculation for one of the combinations, e.g., \(\frac{7}{12} \times \frac{5}{11}\) or \(\frac{5}{12} \times \frac{7}{11}\)
M1: For adding the two correct product probabilities: \(\frac{35}{132} + \frac{35}{132}\)
A1: For the correct simplified fraction: \(\frac{35}{66}\)

First, find the missing frequency \(F\) using the total number of athletes (120):

$$30 + F + 48 + 12 = 120$$

$$90 + F = 120 \implies F = 30$$

Next, calculate the frequency density (FD) for the given class \(30 < t \le 50\):

$$\text{FD} = \frac{\text{Frequency}}{\text{Class Width}} = \frac{48}{20} = 2.4$$

We are given that a frequency density of 2.4 is represented by a bar of height 6 cm. This allows us to find the scale factor:

$$\text{Scale Factor} = \frac{6\text{ cm}}{2.4} = 2.5\text{ cm per unit of FD}$$

Now, calculate the frequency density for the target class \(20 < t \le 30\):

$$\text{FD} = \frac{\text{Frequency}}{\text{Class Width}} = \frac{30}{10} = 3.0$$

Using our scale factor, calculate the height of this bar:

$$\text{Height} = 3.0 \times 2.5 = 7.5\text{ cm}$$

M1: For calculating the missing frequency: \(F = 120 - (30 + 48 + 12) = 30\)
M1: For calculating the frequency density of the \(30 < t \le 50\) class: \(48 \div 20 = 2.4\)
M1: For finding the frequency density of the target class (\(30 \div 10 = 3\)) and using the ratio of heights: \(\text{Height} = 6 \times \frac{3.0}{2.4}\) (or equivalent)
A1: For \(7.5\) (or \(7.5\text{ cm}\))

First, find the value of \(g(3)\):

$$g(3) = 2(3) + 5 = 11$$

We need to find \(x\) such that:

$$f^{-1}(x) = 11$$

By the definition of an inverse function, this is equivalent to:

$$x = f(11)$$

Substitute 11 into the expression for \(f(x)\):

$$f(11) = \frac{3(11)}{11-2} = \frac{33}{9}$$

Simplify the fraction:

$$x = \frac{11}{3}$$

Alternatively, find the expression for \(f^{-1}(x)\):
Let \(y = \frac{3x}{x-2}\)

$$y(x-2) = 3x \implies xy - 2y = 3x$$

$$xy - 3x = 2y \implies x(y-3) = 2y$$

$$x = \frac{2y}{y-3} \implies f^{-1}(x) = \frac{2x}{x-3}$$

Set \(f^{-1}(x) = 11\):

$$\frac{2x}{x-3} = 11 \implies 2x = 11(x-3) \implies 2x = 11x - 33 \implies 9x = 33 \implies x = \frac{11}{3}$$

M1: For evaluating \(g(3) = 11\)
M1: For setting up the equation \(f^{-1}(x) = 11\) or initiating the process to find the inverse function, e.g., \(y(x-2) = 3x\)
M1: For stating \(x = f(11)\) or obtaining the correct inverse function \(f^{-1}(x) = \frac{2x}{x-3}\)
A1: For the final simplified fraction: \(\frac{11}{3}\) (or \(3 \frac{2}{3}\))

Use the formula for the area of a triangle:

$$\text{Area} = \frac{1}{2} b c \sin A$$

$$45 = \frac{1}{2} \times 10 \times 12 \times
sin(\angle BAC)$$

$$45 = 60 \sin(\angle BAC)$$

$$\sin(\angle BAC) = \frac{45}{60} = 0.75$$

Since angle \(BAC\) is obtuse (between \(90^\circ\) and \(180^\circ\)):

$$\angle BAC = 180^\circ - \sin^{-1}(0.75) \approx 180^\circ - 48.59^\circ = 131.41^\circ$$

Now, use the Cosine Rule to find \(BC\):

$$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)$$

$$BC^2 = 10^2 + 12^2 - 2(10)(12)\cos(131.41^\circ)$$

$$BC^2 = 100 + 144 - 240(-0.66144)$$

$$BC^2 = 244 + 158.75 = 402.75$$

$$BC = \sqrt{402.75} \approx 20.068\text{ cm}$$

Rounding to 3 significant figures gives \(BC = 20.1\text{ cm}\).

M1: For using the area formula to find \(\sin(\angle BAC) = 0.75\)
M1: For calculating the obtuse angle: \(\angle BAC \approx 131.4^\circ\) (accept \(131^\circ\) or \(131.41^\circ\))
M1: For substituting their angle into the Cosine Rule: \(BC^2 = 10^2 + 12^2 - 2 \times 10 \times 12 \times \cos(131.4^\circ)\)
A1: For \(20.1\) (accept answers in range 20.0 to 20.1)

Substitute the linear equation \(y = 2x - 1\) into the quadratic equation:

$$x^2 + x(2x - 1) = 10$$

Expand the terms:

$$x^2 + 2x^2 - x = 10$$

Combine like terms to form a 3-term quadratic equation:

$$3x^2 - x - 10 = 0$$

Factorise the quadratic expression:

$$(3x + 5)(x - 2) = 0$$

This gives two solutions for \(x\):

$$x = 2 \quad \text{or} \quad x = -\frac{5}{3}$$

Substitute each \(x\)-value back into the linear equation \(y = 2x - 1\) to find the corresponding \(y\)-values:

When \(x = 2\):

$$y = 2(2) - 1 = 3$$

When \(x = -\frac{5}{3}\):

$$y = 2\left(-\frac{5}{3}\right) - 1 = -\frac{10}{3} - \frac{3}{3} = -\frac{13}{3}$$

So the two sets of solutions are:

$$x = 2, y = 3 \quad \text{and} \quad x = -\frac{5}{3}, y = -\frac{13}{3}$$
(Decimal equivalents: \(x = -1.67, y = -4.33\) are also accepted)

M1: For substituting \(y = 2x - 1\) into the quadratic equation to get a single equation in \(x\)
M1: For simplifying to standard 3-term quadratic form: \(3x^2 - x - 10 = 0\)
M1: For factorising or using the quadratic formula to find both \(x\)-values: \(x = 2\) and \(x = -\frac{5}{3}\) (or \(-1.67\))
A1: For both correct pairs of solutions: \(x = 2, y = 3\) and \(x = -\frac{5}{3}, y = -\frac{13}{3}\) (accept standard decimals rounded to 3sf)

To find where the curve is decreasing, we first find the derivative of \(y\) with respect to \(x\): \(\frac{dy}{dx} = 6x^2 - 18x - 24\). For a decreasing curve, we require \(\frac{dy}{dx} < 0\) (or \(\le 0\)). This gives the quadratic inequality: \(6x^2 - 18x - 24 < 0\). Dividing both sides by 6: \(x^2 - 3x - 4 < 0\). Factoring the quadratic: \((x - 4)(x + 1) < 0\). The critical values are \(x = -1\) and \(x = 4\). Since the quadratic is less than 0, the solution lies between the critical values: \(-1 < x < 4\).

M1: for differentiating at least two terms correctly (e.g., \(6x^2\), \(-18x\), or \(-24\)). M1: for setting the derivative \(< 0\) or \(= 0\) to find critical values (e.g., \(6x^2 - 18x - 24 < 0\) or \(= 0\)). M1: for finding the critical values \(x = -1\) and \(x = 4\). A1: for the correct final inequality: \(-1 < x < 4\) (accept \(-1 \le x \le 4\) or equivalent interval notation).

Let the width of the rectangular base be \(x\text{ cm}\), the length of the base be \(2x\text{ cm}\), and the height of the box be \(y\text{ cm}\).

1. **Express the height \(y\) in terms of \(x\)**:
The volume \(V\) of the box is given by:
\[V = \text{length} \times \text{width} \times \text{height} = 2x \times x \times y = 2x^2 y\]
Given that the volume is \(288\text{ cm}^3\):
\[2x^2 y = 288 \implies y = \frac{144}{x^2}\]

2. **Express the total surface area \(A\) in terms of \(x\)**:
Since the box is open-topped, it has 5 faces:
- The base: \(2x \times x = 2x^2\)
- Two side faces: \(2 \times (2x \times y) = 4xy\)
- Two end faces: \(2 \times (x \times y) = 2xy\)

\[A = 2x^2 + 4xy + 2xy = 2x^2 + 6xy\]
Substitute \(y = \frac{144}{x^2}\) into the surface area equation:
\[A = 2x^2 + 6x\left(\frac{144}{x^2}\right) = 2x^2 + \frac{864}{x}\]

3. **Differentiate \(A\) with respect to \(x\)**:
\[\frac{dA}{dx} = 4x - \frac{864}{x^2}\]

4. **Find the value of \(x\) for which \(A\) is minimum**:
Set \(\frac{dA}{dx} = 0\):
\[4x - \frac{864}{x^2} = 0\]
\[4x^3 = 864\]
\[x^3 = 216\]
\[x = 6\]

5. **Calculate the minimum surface area**:
Substitute \(x = 6\) back into the expression for \(A\):
\[A = 2(6)^2 + \frac{864}{6} = 72 + 144 = 216\text{ cm}^2\]

**M1**: For setting up the volume relation \(2x^2 y = 288\) and expressing \(y\) in terms of \(x\) (e.g., \(y = \frac{144}{x^2}\)).

**M1**: For writing a correct expression for the surface area of the open-topped box in terms of \(x\) only: \(A = 2x^2 + \frac{864}{x}\) (allow slips in algebra during substitution).

**M1**: For differentiating their expression for \(A\) with respect to \(x\) (at least one term differentiated correctly, e.g., \(4x\) or \(-864x^{-2}\)).

**A1**: For setting their derivative to \(0\) and correctly solving to find \(x = 6\).

**A1**: For the correct final answer of \(216\) (or \(216\text{ cm}^2\)).

We are given the equation of the curve \(C\):
\[y = kx^{-1} + x^2 - 9x\]

Differentiating with respect to \(x\):
\[\frac{dy}{dx} = -kx^{-2} + 2x - 9 = -\frac{k}{x^2} + 2x - 9\]

We are given that at \(x = -3\), the gradient of the curve is \(-27\). Substituting these values into the derivative:
\[-\frac{k}{(-3)^2} + 2(-3) - 9 = -27\]
\[-\frac{k}{9} - 6 - 9 = -27\]
\[-\frac{k}{9} - 15 = -27\]
\[-\frac{k}{9} = -12\]
\[k = 108\]

So the equation of the curve is \(y = \frac{108}{x} + x^2 - 9x\), and the derivative is:
\[\frac{dy}{dx} = -\frac{108}{x^2} + 2x - 9\]

At a stationary point, \(\frac{dy}{dx} = 0\):
\[-\frac{108}{x^2} + 2x - 9 = 0\]

Since \(x \neq 0\), we can multiply the entire equation by \(x^2\):
\[-108 + 2x^3 - 9x^2 = 0 \implies 2x^3 - 9x^2 - 108 = 0\]

We can find the integer solutions to this cubic equation by testing factors of \(108\).

For \(x = 6\):
\[2(6)^3 - 9(6)^2 - 108 = 2(216) - 9(36) - 108 = 432 - 324 - 108 = 0\]

So \(x = 6\) is a solution.

Factoring the cubic expression:
\[2x^3 - 9x^2 - 108 = (x - 6)(2x^2 + 3x + 18) = 0\]

The quadratic equation \(2x^2 + 3x + 18 = 0\) has discriminant:
\[\Delta = 3^2 - 4(2)(18) = 9 - 144 = -135 < 0\]

Thus, there are no other real solutions. The only stationary point is at \(x = 6\).

To find the \(y\)-coordinate of the stationary point, substitute \(x = 6\) into the original equation of the curve:
\[y = \frac{108}{6} + 6^2 - 9(6) = 18 + 36 - 54 = 0\]

Therefore, the coordinates of the stationary point are \((6, 0)\).

M1: For differentiating to obtain a correct form of \(\frac{dy}{dx} = -\frac{k}{x^2} + 2x - 9\) (allow one slip).
A1: For substituting \(x = -3\) and gradient \(= -27\) to find \(k = 108\).
M1: For setting their derivative equal to 0, forming the cubic equation \(2x^3 - 9x^2 - 108 = 0\), and finding the root \(x = 6\).
A1: For substituting \(x = 6\) into the original curve equation to find \(y = 0\) and stating the coordinates as \((6, 0)\).

The number of counters of each colour is:
- Red: \(x\)
- Blue: \(2x\)
- Green: \(5\)

The total number of counters in the bag is:
\[N = x + 2x + 5 = 3x + 5\]

When two counters are chosen without replacement, the total number of outcomes is \((3x+5)(3x+4)\).

The probability of selecting two red counters is:
\[P(\text{Red, Red}) = \frac{x(x-1)}{(3x+5)(3x+4)}\]

The probability of selecting two blue counters is:
\[P(\text{Blue, Blue}) = \frac{2x(2x-1)}{(3x+5)(3x+4)}\]

The probability of selecting two green counters is:
\[P(\text{Green, Green}) = \frac{5 \times 4}{(3x+5)(3x+4)} = \frac{20}{(3x+5)(3x+4)}\]

The probability that both counters are of the same colour is given as \(\frac{13}{38}\):
\[\frac{x(x-1) + 2x(2x-1) + 20}{(3x+5)(3x+4)} = \frac{13}{38}\]

Simplify the numerator:
\[x^2 - x + 4x^2 - 2x + 20 = 5x^2 - 3x + 20\]

Simplify the denominator:
\[(3x+5)(3x+4) = 9x^2 + 27x + 20\]

Set up the algebraic equation:
\[\frac{5x^2 - 3x + 20}{9x^2 + 27x + 20} = \frac{13}{38}\]

Cross-multiplying yields:
\[38(5x^2 - 3x + 20) = 13(9x^2 + 27x + 20)\]
\[190x^2 - 114x + 760 = 117x^2 + 351x + 260\]

Rearranging all terms to one side:
\[(190 - 117)x^2 - (114 + 351)x + (760 - 260) = 0\]
\[73x^2 - 465x + 500 = 0\]

Using the quadratic formula to solve for \(x\):
\[x = \frac{465 \pm \sqrt{(-465)^2 - 4(73)(500)}}{2(73)}\]
\[x = \frac{465 \pm \sqrt{216225 - 146000}}{146}\]
\[x = \frac{465 \pm \sqrt{70225}}{146}\]

Since \(\sqrt{70225} = 265\):
\[x = \frac{465 \pm 265}{146}\]

This gives two possible values:
\[x = \frac{730}{146} = 5 \quad \text{or} \quad x = \frac{200}{146} = \frac{100}{73}\]

Since the number of counters \(x\) must be a positive integer, we conclude that \(x = 5\).

M1: For expressing the total number of counters as \(3x + 5\) and writing an algebraic sum of the probabilities of selecting two counters of the same colour, e.g., \(\frac{x(x-1) + 2x(2x-1) + 20}{(3x+5)(3x+4)}\).
M1: For equating this expression to \(\frac{13}{38}\) and showing a correct step in expanding or cross-multiplying.
A1: For forming the correct quadratic equation \(73x^2 - 465x + 500 = 0\) (or equivalent).
A1: For solving the quadratic equation to obtain \(x = 5\) (and rejecting the non-integer solution).

Let the number of yellow beads be \(n - 5\).

Since Sandra selects two beads without replacement, the probability of selecting two yellow beads is:

\(\text{P(Yellow, Yellow)} = \frac{n-5}{n} \times \frac{n-6}{n-1}\)

We are given that this probability is \(\frac{3}{7}\), so:

\(\frac{(n-5)(n-6)}{n(n-1)} = \frac{3}{7}\)

Expand the numerator and the denominator:

\(\frac{n^2 - 11n + 30}{n^2 - n} = \frac{3}{7}\)

Cross-multiply to clear the fractions:

\(7(n^2 - 11n + 30) = 3(n^2 - n)\)

\(7n^2 - 77n + 210 = 3n^2 - 3n\)

Rearrange into a standard quadratic equation form:

\(4n^2 - 74n + 210 = 0\)

Divide the entire equation by 2:

\(2n^2 - 37n + 105 = 0\)

Factorise the quadratic equation:

\((2n - 7)(n - 15) = 0\)

This gives two possible solutions:

\(n = 3.5\) or \(n = 15\)

Since \(n\) must be an integer (and we must have more than 5 beads in total because there are 5 red beads and at least 2 yellow beads), we reject \(n = 3.5\).

Therefore, \(n = 15\).

M1: Writes a correct expression for the probability of selecting two yellow beads without replacement, e.g., \(\frac{n-5}{n} \times \frac{n-6}{n-1}\).
M1: Equates the probability expression to \(\frac{3}{7}\) and expands to form \(\frac{n^2 - 11n + 30}{n^2 - n} = \frac{3}{7}\).
M1: Cross-multiplies and simplifies to obtain a correct 3-term quadratic equation, e.g., \(2n^2 - 37n + 105 = 0\) (or any integer multiple).
M1: Correctly factorises or uses the quadratic formula to solve their quadratic equation, yielding \(n = 15\) (and optionally \(3.5\)).
A1: Correct value of \(n = 15\) only, with the non-integer solution rejected.

The perimeter, \(P\), consists of three sides of the rectangle and the curved boundary of the semicircle:

\(P = 2y + x + \pi \left(\frac{x}{2}\right) = 120\)

Express \(y\) in terms of \(x\):

\(2y = 120 - x - \frac{\pi x}{2}\)

\(y = 60 - \frac{x}{2} - \frac{\pi x}{4}\)

The area, \(A\), of the flower bed is the sum of the area of the rectangle and the area of the semicircle:

\(A = xy + \frac{1}{2}\pi\left(\frac{x}{2}\right)^2 = x\left(60 - \frac{x}{2} - \frac{\pi x}{4}\right) + \frac{\pi x^2}{8}\)

\(A = 60x - \frac{x^2}{2} - \frac{\pi x^2}{4} + \frac{\pi x^2}{8}\)

\(A = 60x - x^2\left(\frac{1}{2} + \frac{\pi}{8}\right)\)

To find the maximum area, differentiate \(A\) with respect to \(x\):

\(\frac{\text{d}A}{\text{d}x} = 60 - 2x\left(\frac{1}{2} + \frac{\pi}{8}\right) = 60 - x\left(1 + \frac{\pi}{4}\right)\)

Set \(\frac{\text{d}A}{\text{d}x} = 0\) for a stationary point:

\(60 - x\left(1 + \frac{\pi}{4}\right) = 0\)

\(x = \frac{60}{1 + \pi/4} = \frac{240}{4+\pi} \approx 33.606\text{ m}\)

Substitute this value of \(x\) back into the area formula:

\(A = 60\left(\frac{240}{4+\pi}\right) - \left(\frac{240}{4+\pi}\right)^2\left(\frac{4+\pi}{8}\right)\)

\(A = \frac{14400}{4+\pi} - \frac{57600(4+\pi)}{8(4+\pi)^2} = \frac{14400}{4+\pi} - \frac{7200}{4+\pi} = \frac{7200}{4+\pi}\)

\(A \approx 1008.18\text{ m}^2\)

Rounding to 3 significant figures gives 1010 \(\text{m}^2\).

M1: Sets up a correct formula for the perimeter, e.g., \(2y + x + \frac{\pi x}{2} = 120\), and rearranges to express \(y\) in terms of \(x\).
M1: Substitutes \(y\) into the area equation to derive a single variable expression for area, e.g., \(A = 60x - x^2\left(\frac{1}{2} + \frac{\pi}{8}\right)\).
M1: Correctly differentiates their area expression to obtain \(\frac{\text{d}A}{\text{d}x} = 60 - x\left(1 + \frac{\pi}{4}\right)\).
M1: Sets \(\frac{\text{d}A}{\text{d}x} = 0\) and solves for \(x\) to find \(x \approx 33.6\) or \(x = \frac{240}{4+\pi}\).
A1: Correct maximum area of 1010 (accept 1008 from rounding to the nearest integer, or 1008.2).

We are given:
\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\)

Since \(OP : PA = 2 : 1\):
\(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\)

Since \(OQ : QB = 3 : 2\):
\(\overrightarrow{OQ} = \frac{3}{5}\mathbf{b}\)

Since the point \(X\) lies on the line \(AQ\), we can write:
\(\overrightarrow{OX} = \overrightarrow{OA} + \lambda \overrightarrow{AQ} = \mathbf{a} + \lambda (\overrightarrow{OQ} - \overrightarrow{OA})\)
\(\overrightarrow{OX} = \mathbf{a} + \lambda \left(\frac{3}{5}\mathbf{b} - \mathbf{a}\right) = (1 - \lambda)\mathbf{a} + \frac{3}{5}\lambda \mathbf{b}\)

Since the point \(X\) also lies on the line \(BP\), we can write:
\(\overrightarrow{OX} = \overrightarrow{OB} + \mu \overrightarrow{BP} = \mathbf{b} + \mu (\overrightarrow{OP} - \overrightarrow{OB})\)
\(\overrightarrow{OX} = \mathbf{b} + \mu \left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\mu \mathbf{a} + (1 - \mu)\mathbf{b}\)

Since \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel vectors, we can equate the coefficients:
1) \(1 - \lambda = \frac{2}{3}\mu\)
2) \ \frac{3}{5}\lambda = 1 - \mu \implies \mu = 1 - \frac{3}{5}\lambda\)

Substitute \(\mu\) into the first equation:
\(1 - \lambda = \frac{2}{3}\left(1 - \frac{3}{5}\lambda\right)\)
\(1 - \lambda = \frac{2}{3} - \frac{2}{5}\lambda\)
\(\frac{1}{3} = \lambda - \frac{2}{5}\lambda\)
\(\frac{1}{3} = \frac{3}{5}\lambda \implies \lambda = \frac{5}{9}\)

Now substitute \(\lambda = \frac{5}{9}\) back into the expression for \(\overrightarrow{OX}\):
\(\overrightarrow{OX} = \left(1 - \frac{5}{9}\right)\mathbf{a} + \frac{3}{5}\left(\frac{5}{9}\right)\mathbf{b} = \frac{4}{9}\mathbf{a} + \frac{1}{3}\mathbf{b}\)

M1: Correctly identifies \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\) and \(\overrightarrow{OQ} = \frac{3}{5}\mathbf{b}\).
M1: Expresses \(\overrightarrow{OX}\) as a linear combination along line \(AQ\), e.g., \((1 - \lambda)\mathbf{a} + \frac{3}{5}\lambda \mathbf{b}\), or along line \(BP\), e.g., \(\frac{2}{3}\mu \mathbf{a} + (1 - \mu)\mathbf{b}\).
M1: Sets up simultaneous equations by equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\).
M1: Solves the simultaneous equations to find the parameter \(\lambda = \frac{5}{9}\) (or \(\mu = \frac{2}{3}\)).
A1: Correct final expression \(\frac{4}{9}\mathbf{a} + \frac{1}{3}\mathbf{b}\) (or equivalent form).