2025 Edexcel IGCSE Mathematics (Specification A) Practice Paper with Answers

The total probability of all outcomes is \(1\).
Therefore, the probability of selecting a blue or yellow counter is:
\(1 - 0.35 = 0.65\)

The ratio of blue counters to yellow counters is \(2 : 3\).
This means there are \(2 + 3 = 5\) equal parts in total for the blue and yellow counters.
The fraction of these counters that are yellow is \(\frac{3}{5}\).

To find the probability that the counter is yellow, we multiply the combined probability of blue or yellow by this fraction:
\(0.65 \times \frac{3}{5} = 0.13 \times 3 = 0.39\).

M1 for a method to find the combined probability of blue and yellow counters, i.e., \(1 - 0.35\) (or \(0.65\))
M1 for a complete method to find the probability of a yellow counter, e.g., \("0.65" \times \frac{3}{2+3}\)
A1 for \(0.39\) (or an equivalent fraction, e.g., \(\frac{39}{100}\))

To factorise the expression fully, find the highest common factor (HCF) of the two terms.
For the coefficients \(12\) and \(18\), the HCF is \(6\).
For the variable \(a\), the highest power common to both \(a^3\) and \(a^2\) is \(a^2\).
For the variable \(b\), the highest power common to both \(b^2\) and \(b^5\) is \(b^2\).

Thus, the HCF of the two terms is \(6a^2b^2\).

Divide each term of the original expression by this HCF to find the terms inside the bracket:
\(\frac{12a^3b^2}{6a^2b^2} = 2a\)
\(\frac{18a^2b^5}{6a^2b^2} = 3b^3\)

Therefore, the fully factorised expression is:
\(6a^2b^2(2a - 3b^3)\).

M1 for finding any common factor containing at least one variable (e.g., \(2ab\), \(6a^2b\), etc.) or a partially factorised expression with a non-HCF factor (e.g., \(3a^2b^2(4a - 6b^3)\))
M1 for a partially factorised expression with \(6a^2b^2\) outside the bracket, or having one incorrect power inside the bracket but HCF correct
A1 for \(6a^2b^2(2a - 3b^3)\)

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

Let the normal price be \(x\).
\(0.85 \times x = 544\)

To find the normal price, divide the sale price by \(0.85\):
\(x = \frac{544}{0.85} = 640\)

The normal price of the laptop is \(\text{£}640\).

M1 for recognizing that \(\text{£}544\) represents \(85\%\) of the normal price, e.g., \(0.85 \times \text{price} = 544\) or \(\frac{544}{85}\)
M1 for a complete method to find the normal price, e.g., \(\frac{544}{0.85}\) or \(\frac{544}{85} \times 100\)
A1 for \(640\)

First, calculate the gradient \(m\) of the line \(L\) using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\):
\(m = \frac{-7 - 5}{4 - (-2)} = \frac{-12}{6} = -2\)

Now, use the equation of a straight line, \(y = mx + c\), with \(m = -2\) and one of the given points, for example \((-2, 5)\):
\(5 = -2(-2) + c\)
\(5 = 4 + c\)
\(c = 1\)

Therefore, the equation of the line \(L\) is:
\(y = -2x + 1\).

M1 for a method to find the gradient, e.g., \(\frac{-7 - 5}{4 - (-2)}\) or a gradient of \(-2\)
M1 for substituting \((-2, 5)\) or \((4, -7)\) into \(y = mx + c\) with their gradient \(m\) to find \(c\)
A1 for \(y = -2x + 1\) (or an equivalent equation in the form \(y = mx + c\))

In the right-angled triangle \(ABC\):
- \(AB\) is the side adjacent to angle \(BAC\).
- \(AC\) is the hypotenuse.

We use the cosine ratio:
\(\cos(BAC) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC}\)

Substitute the given lengths:
\(\cos(BAC) = \frac{7.5}{12.3}\)

Calculate angle \(BAC\):
\(\text{Angle } BAC = \cos^{-1}\left(\frac{7.5}{12.3}\right)\)
\(\text{Angle } BAC \approx \cos^{-1}(0.609756)\)
\(\text{Angle } BAC \approx 52.428^\circ\)

To 1 decimal place, the size of angle \(BAC\) is \(52.4^\circ\).

M1 for identifying and writing the correct trigonometric ratio, e.g., \(\cos(BAC) = \frac{7.5}{12.3}\) or \(\cos(x) = 0.609...\)
M1 for a complete method to find the angle, e.g., \(\cos^{-1}\left(\frac{7.5}{12.3}\right)\)
A1 for \(52.4\) (accept \(52.4^\circ\) or values in the range \(52.4\) to \(52.43\))

First, expand the double brackets \((3x - 4)(2x + 5)\):
\((3x - 4)(2x + 5) = 3x(2x) + 3x(5) - 4(2x) - 4(5)\)
\(= 6x^2 + 15x - 8x - 20\)
\(= 6x^2 + 7x - 20\)

Next, expand the term \(-x(x - 3)\):
\(-x(x - 3) = -x^2 + 3x\)

Now, combine the two expansions:
\((6x^2 + 7x - 20) + (-x^2 + 3x)\)
\(= 6x^2 - x^2 + 7x + 3x - 20\)
\(= 5x^2 + 10x - 20\).

M1 for a correct expansion of the double brackets with at least 3 correct terms out of 4 (e.g., \(6x^2 + 15x - 8x - 20\)) OR a correct expansion of \(-x(x-3)\) to \(-x^2 + 3x\)
M1 for both expansions correct (e.g., \(6x^2 + 7x - 20\) and \(-x^2 + 3x\))
A1 for \(5x^2 + 10x - 20\)

The interest rate is \(2.5\%\) per annum, which corresponds to a multiplier of \(1 + 0.025 = 1.025\).

To find the total value of the investment after 3 years, we use the compound interest formula:
\(\text{Total Value} = 8000 \times (1.025)^3\)
\(= 8000 \times 1.076890625\)
\(= 8615.125\)

To find the total interest earned, subtract the initial investment from the total value:
\(\text{Interest Earned} = 8615.125 - 8000 = 615.125\)

Rounding to the nearest penny (2 decimal places), the interest earned is \(\text{£}615.13\).

M1 for a method to calculate the total value of the investment after 3 years, e.g., \(8000 \times 1.025^3\) (or \(8615.13\))
M1 for subtracting the initial investment of \(8000\) from their total value, or calculating year-on-year interest correctly
A1 for \(615.13\) (do not accept \(8615.13\))

The total number of sweets in the bag is \(6 + 4 = 10\).
Since Maya takes the sweets without replacement, the total number of sweets decreases by 1 for the second pick.

There are two ways to select sweets of different colours:
1. Red first, then Green (RG)
2. Green first, then Red (GR)

Calculate the probability for each path:
- \(P(\text{RG}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}\)
- \(P(\text{GR}) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}\)

Add the two probabilities together to find the total probability of getting different colours:
\(P(\text{different colours}) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90}\)

Simplify the fraction:
\(\frac{48}{90} = \frac{8}{15}\).

M1 for calculating one of the correct probabilities for different colours, e.g., \(\frac{6}{10} \times \frac{4}{9}\) or \(\frac{4}{10} \times \frac{6}{9}\) (or \(\frac{24}{90}\))
M1 for adding both possible different-colour scenarios: \(\frac{6}{10} \times \frac{4}{9} + \frac{4}{10} \times \frac{6}{9}\)
A1 for \(\frac{8}{15}\) (or equivalent fraction, or decimal \(0.533...\) with at least 3 significant figures)

Let the normal price be \(x\). The price is reduced by 15%, so \(0.85x = 357\). Therefore, \(x = \frac{357}{0.85} = 420\).

M1 for setting up the equation \(0.85x = 357\) or indicating that 85% is equivalent to £357 (e.g., \(357 \div 85\)). M1 for a complete method to find the normal price (e.g., \(357 \div 0.85\)). A1 for 420.

First, factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Next, factorise the denominator: \(x^2 - 9 = (x + 3)(x - 3)\). Now, divide the numerator and denominator by the common factor \((x - 3)\): \(\frac{(2x + 1)(x - 3)}{(x + 3)(x - 3)} = \frac{2x + 1}{x + 3}\).

M1 for factorising the quadratic numerator into two brackets, e.g., \((2x + 1)(x - 3)\). M1 for factorising the denominator as a difference of two squares \((x - 3)(x + 3)\). A1 for \(\frac{2x + 1}{x + 3}\).

Let the angle be \(\theta\). The ladder forms a right-angled triangle where the hypotenuse is 6.5 m and the adjacent side to \(\theta\) is 2.5 m. Using the cosine ratio: \(\cos(\theta) = \frac{2.5}{6.5}\). Therefore, \(\theta = \cos^{-1}\left(\frac{2.5}{6.5}\right) \approx 67.38^\circ\). Rounding to 1 decimal place gives \(67.4^\circ\).

M1 for setting up the trigonometric ratio, e.g., \(\cos(\theta) = \frac{2.5}{6.5}\). M1 for \(\cos^{-1}\left(\frac{2.5}{6.5}\right)\). A1 for 67.4.

The total number of counters is \(5 + 3 = 8\). The probability of choosing two red counters is: \(P(\text{Red, Red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}\). The probability of choosing two blue counters is: \(P(\text{Blue, Blue}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}\). The probability that both are the same colour is: \(P(\text{same colour}) = \frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}\).

M1 for finding one correct product of probabilities, e.g., \(\frac{5}{8} \times \frac{4}{7}\) or \(\frac{3}{8} \times \frac{2}{7}\). M1 for adding the two correct products. A1 for \(\frac{13}{28}\) (or equivalent fraction or decimal \\approx 0.464).

The gradient of the given line is \(m_1 = 3\). Since the lines are perpendicular, the gradient of the new line is \(m_2 = -\frac{1}{3}\). Using the equation of a straight line with the point \((6, 2)\): \(y - 2 = -\frac{1}{3}(x - 6)\) which simplifies to \(y - 2 = -\frac{1}{3}x + 2\), or \(y = -\frac{1}{3}x + 4\).

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

The formula for compound interest is \(A = P(1 + r)^t\). Substituting the given values: \(A = 8000 \times (1.025)^3 = 8615.125\). The total interest earned is the final amount minus the initial investment: \(\text{Interest} = 8615.125 - 8000 = 615.125\). Rounding to the nearest penny gives £615.13.

M1 for setting up the compound interest calculation, e.g., \(8000 \times 1.025^3\). M1 for finding the total amount \(8615.125\) or subtracting 8000 from their total amount. A1 for 615.13.

Multiply every term by \(x(x + 1)\) to clear the fractions: \(12(x + 1) - 12x = x(x + 1)\). Expand both sides: \(12x + 12 - 12x = x^2 + x\), which simplifies to \(12 = x^2 + x\). Rearrange into a quadratic equation: \(x^2 + x - 12 = 0\). Factorise the quadratic: \((x + 4)(x - 3) = 0\). Therefore, \(x = -4\) or \(x = 3\).

M1 for clearing fractions correctly to get \(12(x + 1) - 12x = x(x + 1)\). M1 for writing as a three-term quadratic equation, e.g., \(x^2 + x - 12 = 0\). A1 for both correct solutions \(x = 3\) and \(x = -4\).

Let \(x\) be the number of students who study both languages. The number of students studying at least one language is \(30 - 5 = 25\). Therefore: \((18 - x) + x + (15 - x) = 25\). This simplifies to \(33 - x = 25\), so \(x = 8\). The probability that a randomly chosen student studies both is \(\frac{8}{30} = \frac{4}{15}\).

M1 for a correct method to find the number of students studying both languages, e.g., \(18 + 15 - (30 - 5)\). M1 for finding the number of students who study both is 8. A1 for \(\frac{4}{15}\) (or equivalent fraction or decimal \\approx 0.267).

(a) First, expand \((3x - 5)(2x + 3)\):
\((3x - 5)(2x + 3) = 6x^2 + 9x - 10x - 15 = 6x^2 - x - 15\)

Next, expand \((x - 4)^2\):
\((x - 4)^2 = x^2 - 8x + 16\)

Now subtract the second expression from the first:
\((6x^2 - x - 15) - (x^2 - 8x + 16) = 6x^2 - x - 15 - x^2 + 8x - 16 = 5x^2 + 7x - 31\)

(b) Expand and rearrange the equation:
\(x^2 - x - 6 = 14\)

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

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

Thus, \(x = 5\) or \(x = -4\).

(a)
M1: for expanding at least one of the expressions correctly, e.g., \(6x^2 - x - 15\) or \(x^2 - 8x + 16\) (allow one sign error)
M1: for subtracting the quadratic terms correctly, e.g., showing \(6x^2 - x^2\) and \(-15 - 16\)
A1: for the final fully simplified expression \(5x^2 + 7x - 31\)

(b)
M1: for expanding LHS and rearranging to form a three-term quadratic equal to zero, i.e., \(x^2 - x - 20 = 0\)
A1: for both correct solutions \(x = 5\) and \(x = -4\)

(a) The probability of picking either a blue or a green marble is \(1 - 0.3 = 0.7\).
The ratio of blue to green marbles is \(3:4\), so the probability is split into \(3 + 4 = 7\) parts.
Each part is worth \(\frac{0.7}{7} = 0.1\).
Therefore, the probability of picking a blue marble is \(3 \times 0.1 = 0.3\).

(b) The probability of picking a green marble is \(4 \times 0.1 = 0.4\).
Let \(N\) be the total number of marbles in the bag. Since there are 24 green marbles, we can write:
\(0.4 \times N = 24\)

\(N = \frac{24}{0.4} = 60\).

(a)
M1: for finding the combined probability of blue and green, \(1 - 0.3 = 0.7\)
M1: for dividing the combined probability by the sum of ratio parts, e.g., \(\frac{0.7}{7} \times 3\)
A1: for \(0.3\) (or equivalent fraction, e.g. \(\frac{3}{10}\))

(b)
M1: for establishing the probability of green is \(0.4\) and writing an equation such as \(0.4N = 24\)
A1: for \(60\)

(a) Using Pythagoras' Theorem, \(AB^2 + BC^2 = AC^2\):
\((x + 2)^2 + (2x - 3)^2 = (2x + 1)^2\)

Expand each bracket:
\((x^2 + 4x + 4) + (4x^2 - 12x + 9) = 4x^2 + 4x + 1\)

Simplify the left-hand side:
\(5x^2 - 8x + 13 = 4x^2 + 4x + 1\)

Subtract \(4x^2 + 4x + 1\) from both sides:
\(x^2 - 12x + 12 = 0\) (as required).

(b) Solve \(x^2 - 12x + 12 = 0\) using the quadratic formula:
\(x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(12)}}{2(1)} = \frac{12 \pm \sqrt{96}}{2}\)

This gives two values:
\(x = 6 + 2\sqrt{6} \approx 10.899\)
\(x = 6 - 2\sqrt{6} \approx 1.101\)

Since the side \(BC = 2x - 3\) must be positive, \(x\) must be greater than \(1.5\). Therefore, we discard \(1.101\).
Using \(x \approx 10.899\):
\(AC = 2(10.899) + 1 \approx 22.798\text{ cm}\)

To 3 significant figures, \(AC = 22.8\text{ cm}\).

(a)
M1: for substituting the algebraic terms into Pythagoras' Theorem, i.e., \((x + 2)^2 + (2x - 3)^2 = (2x + 1)^2\)
M1: for expanding the brackets correctly, showing at least two of the three expanded quadratics correctly
A1: for fully simplifying and obtaining the given equation \(x^2 - 12x + 12 = 0\) with no algebraic errors

(b)
M1: for solving the quadratic equation to find the valid value of \(x\) (accept \(x = 6 + 2\sqrt{6}\) or \(x \approx 10.9\))
A1: for \(22.8\) (accept answers in range \(22.79 - 22.8\))

(a) Let the original price of the television be \(P\).
After the first reduction of \(15\%\), the price is \(0.85P\).
After the second reduction of \(10\%\), the price is:
\(0.90 \times 0.85P = 0.765P\)

The single multiplier is \(0.765\).
This corresponds to a percentage reduction of:
\((1 - 0.765) \times 100\% = 23.5\%\).

(b) Let the original price be \(P\).
We have:
\(0.765P = 382.50\)

\(P = \frac{382.50}{0.765} = 500\)

Thus, the original price of the television was \(\text{£}500\).

(a)
M1: for calculating the combined multiplier \(0.85 \times 0.90\) (or equivalent percentage calculations, e.g. starting with £100)
A1: for \(23.5\%\) (or equivalent)

(b)
M1: for a correct algebraic setup, e.g., \(0.765P = 382.5\) or \(\text{New Price} \div 0.9 \div 0.85\)
M1: for evaluating \(\frac{382.5}{0.765}\) or \(\frac{382.5}{0.9} = 425\) then \(\frac{425}{0.85}\)
A1: for \(500\) (or \(\text{£}500\))

(a) Differentiating \(y = 2x^3 - 9x^2 - 24x + 7\) term by term with respect to \(x\):
\(\frac{\text{d}y}{\text{d}x} = 6x^2 - 18x - 24\)

(b) At the turning points, \(\frac{\text{d}y}{\text{d}x} = 0\):
\(6x^2 - 18x - 24 = 0\)

Divide by 6:
\(x^2 - 3x - 4 = 0\)

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

This gives \(x = 4\) and \(x = -1\).

Find the corresponding \(y\)-coordinates:
For \(x = 4\):
\(y = 2(4)^3 - 9(4)^2 - 24(4) + 7 = 128 - 144 - 96 + 7 = -105\)

For \(x = -1\):
\(y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 7 = -2 - 9 + 24 + 7 = 20\)

So the coordinates of the turning points are \((4, -105)\) and \((-1, 20)\).

(a)
M1: for differentiating at least two terms correctly
A1: for \(6x^2 - 18x - 24\)

(b)
M1: for setting their \(\frac{\text{d}y}{\text{d}x} = 0\) and solving the quadratic equation to find two values of \(x\) (e.g. \(x = 4, x = -1\))
M1: for substituting at least one \(x\)-value back into the original curve equation to find a \(y\)-value
A1: for both correct pairs of coordinates \((4, -105)\) and \((-1, 20)\)

(a) To solve:
\(3x - 2y = 6\) (1)
\(2x + 3y = 43\) (2)

Multiply (1) by 3 and (2) by 2:
\(9x - 6y = 18\) (3)

\(4x + 6y = 86\) (4)

Add equations (3) and (4):
\(13x = 104 \Rightarrow x = 8\)

Substitute \(x = 8\) into (1):
\(3(8) - 2y = 6 \Rightarrow 24 - 2y = 6 \Rightarrow 2y = 18 \Rightarrow y = 9\)

So, \(x = 8\) and \(y = 9\).

(b) By substitution, let \(x = 2^a\) and \(y = 3^b\).
From part (a):
\(2^a = 8 \Rightarrow a = 3\)

\(3^b = 9 \Rightarrow b = 2\).

(a)
M1: for a correct method to eliminate one variable, e.g. multiplying equations to match coefficients and adding or subtracting them
M1: for substituting their found variable back to find the other variable
A1: for \(x = 8\) and \(y = 9\)

(b)
M1: for setting \(2^a = 8\) and \(3^b = 9\) (or equivalent)
A1: for both \(a = 3\) and \(b = 2\)

(a) The probability of picking a green counter first is \(\frac{n}{9\)}.
Since the selection is made without replacement, the probability of picking a green counter second is \(\frac{n - 1}{8}\).

The probability that both counters are green is:
\(P(\text{Green, Green}) = \frac{n}{9} \times \frac{n - 1}{8} = \frac{n(n - 1)}{72} = \frac{n^2 - n}{72}\).

(b) Set the probability expression equal to \(\frac{5}{12}\):
\(\frac{n^2 - n}{72} = \frac{5}{12}\)

Multiply both sides by 72:
\(n^2 - n = \frac{5}{12} \times 72 = 30\)

Rearrange into a quadratic equation:
\(n^2 - n - 30 = 0\)

Factorise:
\((n - 6)(n + 5) = 0\)

Since the number of counters \(n\) must be positive, \(n = 6\).

(a)
M1: for writing the product of the two individual probabilities, i.e., \(\frac{n}{9} \times \frac{n-1}{8}\)
A1: for showing the correct algebraic simplification to reach the given result

(b)
M1: for setting up the equation \(\frac{n^2 - n}{72} = \frac{5}{12}\) and rearranging it to a three-term quadratic, i.e., \(n^2 - n - 30 = 0\)
M1: for factorising or solving their quadratic equation to find the positive integer value
A1: for \(6\) (must reject \(n = -5\))

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

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

Now, simplify the fraction:
\(\frac{(2x - 1)(x + 3)}{(x - 3)(x + 3)} = \frac{2x - 1}{x - 3}\)

(b) Find a common denominator, which is \((y-2)(y+1)\):
\(\frac{3}{y-2} - \frac{2}{y+1} = \frac{3(y+1) - 2(y-2)}{(y-2)(y+1)}\)

Expand the terms in the numerator:
\(3(y+1) - 2(y-2) = 3y + 3 - 2y + 4 = y + 7\)

So the simplified single fraction is:
\(\frac{y + 7}{(y-2)(y+1)}\).

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

(b)
M1: for expressing both fractions over a common denominator, e.g., \(\frac{3(y+1) - 2(y-2)}{(y-2)(y+1)}\)
A1: for \(\frac{y+7}{(y-2)(y+1)}\) or \(\frac{y+7}{y^2 - y - 2}\)

We can express \(\overrightarrow{OX}\) in two different ways using two vector paths.

First, we find the position vectors of \(P\) and \(Q\):
Since \(OP : PA = 2 : 1\), \(P\) is \(\frac{2}{3}\) of the way along \(OA\), so:
\(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\)

Since \(OQ : QB = 1 : 3\), \(Q\) is \(\frac{1}{4}\) of the way along \(OB\), so:
\(\overrightarrow{OQ} = \frac{1}{4}\mathbf{b}\)

Now, let's find the vector \(\overrightarrow{AQ}\):
\(\overrightarrow{AQ} = \overrightarrow{AO} + \overrightarrow{OQ} = -\mathbf{a} +
\frac{1}{4}\mathbf{b}\)

Since \(X\) lies on the line \(AQ\), we can write \(\overrightarrow{AX} = \mu \overrightarrow{AQ}\) for some scalar \(\mu\). Therefore:
\(\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = \mathbf{a} + \mu\left(-\mathbf{a} + \frac{1}{4}\mathbf{b}\right) = (1-\mu)\mathbf{a} +
\frac{1}{4}\mu\mathbf{b}\)

Similarly, let's find the vector \(\overrightarrow{BP}\):
\(\overrightarrow{BP} = \overrightarrow{BO} + \overrightarrow{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a}\)

Since \(X\) lies on the line \(BP\), we can write \(\overrightarrow{BX} = k \overrightarrow{BP}\) for some scalar \(k\). Therefore:
\(\overrightarrow{OX} = \overrightarrow{OB} + \overrightarrow{BX} = \mathbf{b} + k\left(-\mathbf{b} + \frac{2}{3}\mathbf{a}\right) = \frac{2}{3}k\mathbf{a} + (1-k)\mathbf{b}\)

Since \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel vectors, we can equate their coefficients from both expressions for \(\overrightarrow{OX}\):

1) For \(\mathbf{a}\): \(1-\mu = \frac{2}{3}k\)
2) For \(\mathbf{b}\): \(\frac{1}{4}\mu = 1-k\)

From equation (2), we get \(k = 1 - \frac{1}{4}\mu\). Substituting this into equation (1):
\(1-\mu = \frac{2}{3}\left(1 - \frac{1}{4}\mu\right)\)
\(1-\mu = \frac{2}{3} - \frac{1}{6}\mu\)

Multiply the entire equation by 6 to clear fractions:
\(6 - 6\mu = 4 - \mu\)
\(2 = 5\mu \implies \mu = \frac{2}{5}\)

Now substitute \(mu = \frac{2}{5}\) back into our first expression for \(\overrightarrow{OX}\):
\(\overrightarrow{OX} = \left(1-\frac{2}{5}\right)\mathbf{a} + \frac{1}{4}\left(\frac{2}{5}\right)\mathbf{b} = \frac{3}{5}\mathbf{a} + \frac{1}{10}\mathbf{b}\)

M1: For finding \(\overrightarrow{OP} =
\frac{2}{3}\mathbf{a}\) or \(\overrightarrow{OQ} =
\frac{1}{4}\mathbf{b}\).
M1: For writing a correct expression for \(\overrightarrow{AQ}\) or \(\overrightarrow{BP}\), e.g., \(\overrightarrow{AQ} = -\mathbf{a} +
\frac{1}{4}\mathbf{b}\).
M1: For expressing \(\overrightarrow{OX}\) in terms of an unknown parameter along either line, e.g., \(\overrightarrow{OX} = (1-\mu)\mathbf{a} +
\frac{1}{4}\mu\mathbf{b}\) or \(\overrightarrow{OX} = \frac{2}{3}k\mathbf{a} + (1-k)\mathbf{b}\).
M1: For setting up simultaneous equations by equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\): \(1-\mu = \frac{2}{3}k\) and \(\frac{1}{4}\mu = 1-k\).
M1: For solving the system to find \(\mu = \frac{2}{5}\) or \(k = \frac{9}{10}\).
A1: For substituting the value of \(\mu\) or \(k\) back into the expression to show \(\overrightarrow{OX} = \frac{3}{5}\mathbf{a} + \frac{1}{10}\mathbf{b}\) with clear working.

The total volume \(V\) of the solid is the sum of the volume of the cylinder and the hemisphere:
\(V = \pi r^2 h + \frac{2}{3}\pi r^3\)

We are given that \(V = 300\pi\), so:
\(\pi r^2 h + \frac{2}{3}\pi r^3 = 300\pi\)

Divide every term by \(\pi\):
\(r^2 h + \frac{2}{3}r^3 = 300\)

Rearrange this equation to make \(h\) the subject:
\(r^2 h = 300 - \frac{2}{3}r^3\)
\(h = \frac{300}{r^2} - \frac{2}{3}r\)

The total surface area \(S\) of the solid is made up of:
1. The circular flat base of the cylinder: \(\pi r^2\)
2. The curved surface area of the cylinder: \(2\pi r h\)
3. The curved surface area of the hemisphere: \(2\pi r^2\)

Summing these areas gives:
\(S = \pi r^2 + 2\pi r h + 2\pi r^2 = 3\pi r^2 + 2\pi r h\)

Substitute our expression for \(h\) into the surface area equation:
\(S = 3\pi r^2 + 2\pi r\left(\frac{300}{r^2} - \frac{2}{3}r\right)\)
\(S = 3\pi r^2 + \frac{600\pi}{r} - \frac{4}{3}\pi r^2\)

Combine the \(\pi r^2\) terms:
\(S = \left(3 - \frac{4}{3}\right)\pi r^2 + \frac{600\pi}{r}\)
\(S = \frac{5}{3}\pi r^2 + \frac{600\pi}{r\)}

M1: For writing a correct expression for the total volume: \(V = \pi r^2 h + \frac{2}{3}\pi r^3\).
M1: For equating the volume to \(300\pi\) and attempting to solve for \(h\).
A1: For obtaining a correct expression for \(h\), e.g., \(h = \frac{300}{r^2} - \frac{2}{3}r\).
M1: For writing a correct expression for the total surface area of the combined solid: \(S = \pi r^2 + 2\pi r h + 2\pi r^2\) (or simplified to \(3\pi r^2 + 2\pi r h\)).
M1: For substituting their expression for \(h\) into their expression for \(S\).
A1: For obtaining the given result \(S = \frac{5}{3}\pi r^2 + \frac{600\pi}{r}\) with full, clear algebraic steps shown.

The multiplier for the first year is \(1.12\). The multiplier for the second year is \(0.95\). The multiplier for the third year is \(1.08\). The overall equation is \(P \times 1.12 \times 0.95 \times 1.08 = 6894.72\). Evaluating the product of the multipliers: \(1.12 \times 0.95 \times 1.08 = 1.14912\). Finding the original value: \(P = 6894.72 \div 1.14912 = 6000\).

M1 for \(1.12 \times 0.95 \times 1.08\) or \(1.14912\). M1 for \(6894.72 \div 1.14912\). A1 for 6000.

Factorising the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Factorising the denominator: \(x^2 - 9 = (x - 3)(x + 3)\). Simplifying the fraction by dividing the numerator and the denominator by the common factor \((x - 3)\): \(\frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} = \frac{2x + 1}{x + 3}\).

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

Since \(BD : DC = 1 : 2\) and \(BC = 12\text{ cm}\), we find \(BD = \frac{1}{3} \times 12 = 4\text{ cm}\). In right-angled triangle \(ABD\), we can use the tangent ratio: \(\tan(\angle ADB) = \frac{AB}{BD} = \frac{7}{4}\). Thus, \(\angle ADB = \tan^{-1}(1.75) \approx 60.255^\circ\). Since \(BDC\) is a straight line, angle \(ADC = 180^\circ - \angle ADB = 180^\circ - 60.255^\circ = 119.745^\circ\). Correct to 1 decimal place, this is \(119.7^\circ\).

M1 for finding \(BD = 4\text{ cm}\). M1 for finding \(\angle ADB = \tan^{-1}(7/4) \approx 60.3^\circ\) or using the cosine rule on triangle \(ADC\). A1 for 119.7.

The total number of counters is \(5 + 3 = 8\). The probability of choosing a red counter and then a blue counter is \(P(R, B) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\). The probability of choosing a blue counter and then a red counter is \(P(B, R) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\). The probability that the two counters are of different colours is the sum of these probabilities: \(P(\text{different}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\).

M1 for \(\frac{5}{8} \times \frac{3}{7}\) or \(\frac{3}{8} \times \frac{5}{7}\). M1 for adding both probabilities: \(\frac{15}{56} + \frac{15}{56}\). A1 for \(\frac{15}{28}\) or equivalent decimal.

The gradient of \(L_1\) is \(m_1 = \frac{8 - 5}{4 - (-2)} = \frac{3}{6} = \frac{1}{2}\). Since \(L_2\) is perpendicular to \(L_1\), its gradient is the negative reciprocal: \(m_2 = -2\). Using the point-slope form with point \((1, -3)\): \(y - (-3) = -2(x - 1)\), which simplifies to \(y + 3 = -2x + 2\), and rearranging gives \(y + 2x = -1\).

M1 for finding the gradient of \(L_1\) as \(\frac{1}{2}\) or the gradient of \(L_2\) as \(-2\). M1 for substituting \((1, -3)\) and gradient \(-2\) into a linear equation. A1 for \(y + 2x = -1\) or any equivalent integer form.

Let \(A\) be the cost of an adult ticket and \(C\) be the cost of a child ticket. We have the simultaneous equations: (1) \(3A + 4C = 58.50\) and (2) \(5A + 2C = 69.50\). Multiplying equation (2) by 2 gives (3) \(10A + 4C = 139.00\). Subtracting equation (1) from (3) gives \(7A = 80.50\). Dividing by 7 gives \(A = 11.50\).

M1 for setting up two correct simultaneous equations. M1 for a correct method to eliminate \(C\). A1 for 11.50 (or 11.5).

Multiply both sides by \(x + 2\) to get \(y(x + 2) = 4 - 3x\). Expanding the left side yields \(xy + 2y = 4 - 3x\). Collect all terms involving \(x\) on one side: \(xy + 3x = 4 - 2y\). Factorise out \(x\) to get \(x(y + 3) = 4 - 2y\). Finally, divide by \(y + 3\) to get \(x = \frac{4 - 2y}{y + 3}\).

M1 for clearing the fraction to get \(y(x+2) = 4-3x\). M1 for collecting terms and factorising to get \(x(y+3) = 4-2y\). A1 for \(x = \frac{4-2y}{y+3}\) or equivalent.

Let \(P\) be the original price. The price after a \(15\%\) increase is \(1.15P\). A \(20\%\) reduction on this price gives \(0.80 \times 1.15P = 0.92P\). We are given that \(0.92P = 73.60\), so dividing by \(0.92\) gives \(P = 73.60 \div 0.92 = 80\).

M1 for finding the overall multiplier \(1.15 \times 0.80 = 0.92\) or expressing \(73.60 \div 0.80 = 92\). M1 for \(73.60 \div 0.92\) or \(92 \div 1.15\). A1 for 80.

Let the original value of the watch be \(V\). After the first year, the value is \(V \times 1.08\). After the second year, the value is \(V \times 1.08 \times 0.95 = 1.026V\). We are given that \(1.026V = 3693.60\), so \(V = \frac{3693.60}{1.026} = 3600\). So the original value of the watch was £3600.

M1: for multiplying multipliers, e.g., \(1.08 \times 0.95\) (or equivalent ratio method). M1: for setting up the equation \(1.026V = 3693.60\) or evaluating \(\frac{3693.60}{1.08 \times 0.95}\). A1: for 3600.

Using the Cosine Rule: \(BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(BAC)\). Substitute the given values: \(BC^2 = 8.4^2 + 6.5^2 - 2 \times 8.4 \times 6.5 \times \cos(112^\circ)\). \(BC^2 = 70.56 + 42.25 - 109.2 \times (-0.3746...)\). \(BC^2 = 112.81 - (-40.907...)\). \(BC^2 = 153.717...\). \(BC = \sqrt{153.717...} \approx 12.398\text{ cm}\). To 3 significant figures, \(BC = 12.4\text{ cm}\).

M1: for substitution into the Cosine Rule formula, e.g., \(8.4^2 + 6.5^2 - 2 \times 8.4 \times 6.5 \times \cos(112^\circ)\). M1: for a correct evaluation of \(BC^2\) (e.g., 153.7...). A1: for 12.4 (accept 12.39 to 12.41).

First, factorise the numerator: \(3x^2 - 12 = 3(x^2 - 4) = 3(x-2)(x+2)\). Next, factorise the denominator: \(x^2 + x - 6 = (x+3)(x-2)\). Now, simplify the fraction by dividing out the common factor \((x-2)\): \(\frac{3(x-2)(x+2)}{(x+3)(x-2)} = \frac{3(x+2)}{x+3}\).

M1: for factorising the numerator to \(3(x^2-4)\) or \(3(x-2)(x+2)\). M1: for factorising the denominator to \((x+3)(x-2)\). A1: for \(\frac{3(x+2)}{x+3}\) or \(\frac{3x+6}{x+3}\).

The total number of counters is \(7 + 5 = 12\). Probability of selecting two red counters: \(P(\text{Red, Red}) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132}\). Probability of selecting two blue counters: \(P(\text{Blue, Blue}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}\). Probability that both counters are the same colour is: \(P(\text{Same Colour}) = P(\text{Red, Red}) + P(\text{Blue, Blue}) = \frac{42}{132} + \frac{20}{132} = \frac{62}{132} = \frac{31}{66}\).

M1: for one correct probability product, e.g., \(\frac{7}{12} \times \frac{6}{11}\) or \(\frac{5}{12} \times \frac{4}{11}\). M1: for adding the two correct probability products, e.g., \(\frac{42}{132} + \frac{20}{132}\). A1: for \(\frac{31}{66}\) (or equivalent fraction, or 0.470 or 0.469...).

Rearrange the equation of \(L_1\) to find its gradient: \(2y = 3x + 8 \implies y = \frac{3}{2}x + 4\). So the gradient of \(L_1\) is \(m_1 = \frac{3}{2}\). Since \(L_2\) is perpendicular to \(L_1\), its gradient \(m_2\) is \(m_2 = -\frac{1}{m_1} = -\frac{2}{3}\). Using the point-slope form with point \((6, -1)\): \(y - (-1) = -\frac{2}{3}(x - 6)\) which simplifies to \(y + 1 = -\frac{2}{3}x + 4\), giving \(y = -\frac{2}{3}x + 3\).

M1: for finding the gradient of \(L_1\) is \(\frac{3}{2}\) (or 1.5). M1: for using the perpendicular gradient rule to find \(m_2 = -\frac{2}{3}\) and substituting \((6, -1)\) into a linear equation. A1: for \(y = -\frac{2}{3}x + 3\) (or equivalent in the requested form, e.g. \(y = -0.667x + 3\)).

First, calculate the frequency density for the class interval \(4 < w \le 10\): \(\text{Class Width} = 10 - 4 = 6\), so \(\text{Frequency Density} = \frac{24}{6} = 4\). Since the height of the bar for this class is 8 cm, the scale of the histogram is \(\frac{8\text{ cm}}{4} = 2\text{ cm per unit of frequency density}\). Now, calculate the frequency density for the class interval \(10 < w \le 15\): \(\text{Class Width} = 15 - 10 = 5\), so \(\text{Frequency Density} = \frac{15}{5} = 3\). Finally, calculate the height of the bar: \(\text{Height} = 3 \times 2 = 6\text{ cm}\).

M1: for finding the frequency density of the \(4 < w \le 10\) class, which is 4 (or establishing a ratio of \(\frac{24}{6}\)). M1: for finding the frequency density of the \(10 < w \le 15\) class, which is 3 (or \(\frac{15}{5}\)). A1: for 6 (or 6 cm).

Multiply both sides by \((2 - x)\): \(y(2 - x) = 3x + 5\). Expand the left-hand side: \(2y - xy = 3x + 5\). Rearrange to bring all terms containing \(x\) to one side: \(2y - 5 = 3x + xy\). Factorise \(x\) on the right-hand side: \(2y - 5 = x(3 + y)\). Divide by \((3 + y)\) to make \(x\) the subject: \(x = \frac{2y - 5}{3 + y}\).

M1: for multiplying both sides by \(2 - x\) and expanding, e.g., \(2y - xy = 3x + 5\). M1: for grouping \(x\) terms on one side and factorising, e.g., \(x(3+y) = 2y-5\). A1: for \(x = \frac{2y - 5}{y + 3}\) (or equivalent, such as \(x = \frac{5 - 2y}{-y - 3}\)).

First, find the slant height \(l\) of the cone using Pythagoras' Theorem: \(l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}\). The total surface area of a cone is given by the formula: \(A = \pi r^2 + \pi r l\). Substitute the values for \(r = 5\) and \(l = 13\): \(A = \pi (5^2) + \pi (5)(13) = 25\pi + 65\pi = 90\pi\text{ cm}^2\).

M1: for using Pythagoras' theorem to find the slant height \(l = 13\). M1: for substitution of \(r = 5\) and their \(l\) into the total surface area formula \(\pi r^2 + \pi r l\). A1: for \(90\pi\) (or 90).

(a) Factorise the numerator: \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\).
Factorise the denominator: \(x^2 - 9 = (x - 3)(x + 3)\).
Thus, \(\frac{(2x - 1)(x + 3)}{(x - 3)(x + 3)} = \frac{2x - 1}{x - 3}\).

(b) Multiply all terms by 12:
\(3(3y - 2) - 4(y - 1) > 8\)
\(9y - 6 - 4y + 4 > 8\)
\(5y - 2 > 8\)
\(5y > 10\)
\(y > 2\)

(a)
M1: For factoring the numerator as \((2x - 1)(x + 3)\) or the denominator as \((x - 3)(x + 3)\).
M1: For both numerator and denominator correctly factorised.
A1: Correct simplified fraction \(\frac{2x - 1}{x - 3}\).

(b)
M1: For a correct method to clear fractions, e.g., multiplying all terms by 12 to get \(3(3y - 2) - 4(y - 1) > 8\) (allow one sign error in expansion).
A1: Correct final inequality \(y > 2\).

(a) Let the original price be \(P\).
At the end of year 1, the value is \(1.15 \times P\).
At the end of year 2, the value is \(1.15 \times 0.92 \times P = 1.058 \times P\).
Given \(1.058 \times P = 908.50\),
\(P = \frac{908.50}{1.058} = 850\).

(b) The compound interest formula gives:
\(1200 \times \left(1 + \frac{r}{100}\right)^3 = 1323\)
\(\left(1 + \frac{r}{100}\right)^3 = \frac{1323}{1200} = 1.1025\)
\(1 + \frac{r}{100} = \sqrt[3]{1.1025} \approx 1.0331003\)
\(\frac{r}{100} = 0.0331003 \implies r \approx 3.31\%\).

(a)
M1: For writing a correct equation or expression for the compound change, e.g., \(P \times 1.15 \times 0.92 = 908.50\) or showing \(1.15 \times 0.92 = 1.058\).
M1: For a complete method to find the original price, e.g., \(\frac{908.50}{1.058}\).
A1: \(850\).

(b)
M1: For setting up the equation \(1200 \times k^3 = 1323\) (where \(k = 1 + r/100\)) or for \(\sqrt[3]{\frac{1323}{1200}}\).
A1: \(3.31\) (accept \(3.31\%\)).

(a) In the right-angled triangle \(ABC\), using Pythagoras' theorem:
\(AC = \sqrt{AB^2 + BC^2} = \sqrt{5^2 + 8^2} = \sqrt{89} \approx 9.43\text{ cm}\).

(b) The projection of the line \(AF\) onto the horizontal base \(BCFE\) is \(BF\).
Since \(BCFE\) is a rectangle, \(\angle BCF = 90^\circ\).
In the right-angled triangle \(BCF\):
\(BF^2 = BC^2 + CF^2 = 8^2 + 12^2 = 208\).
Since the line \(AB\) is vertical, it is perpendicular to the base, so \(\angle ABF = 90^\circ\).
In the right-angled triangle \(ABF\), the required angle \(\theta = \angle AFB\) is given by:
\(\tan(\theta) = \frac{AB}{BF} = \frac{5}{\sqrt{208}}\)
\(\theta = \tan^{-1}\left(\frac{5}{\sqrt{208}}\right) \approx 19.1^\circ\).

(a)
M1: For \(5^2 + 8^2\) or \(\sqrt{5^2 + 8^2}\).
A1: \(9.43\) (accept answers in the range \(9.43 - 9.44\)).

(b)
M1: For finding \(BF^2 = 8^2 + 12^2\) (\(208\)) or \(BF = \sqrt{208}\) (\(\approx 14.4\)).
M1: For using \(\tan(\theta) = \frac{5}{BF}\) or finding \(AF = \sqrt{5^2 + 208} = \sqrt{233}\) (\(\approx 15.26\)) and using \(\sin(\theta) = \frac{5}{AF}\) or \(\cos(\theta) = \frac{BF}{AF}\).
A1: \(19.1\) (accept answers in the range \(19.1 - 19.2\)).

(a) Midpoint \(M = \left(\frac{-2 + 6}{2}, \frac{7 + (-3)}{2}\right) = (2, 2)\).

(b) The gradient of \(PQ\) is:
\(m_{PQ} = \frac{-3 - 7}{6 - (-2)} = \frac{-10}{8} = -\frac{5}{4}\).
Since the required line is perpendicular to \(PQ\), its gradient \(m\) is:
\(m = -\frac{1}{m_{PQ}} = \frac{4}{5}\).
Using the midpoint \((2, 2)\) in the equation of the line:
\(y - 2 = \frac{4}{5}(x - 2)\)
\(5(y - 2) = 4(x - 2)\)
\(5y - 10 = 4x - 8\)
\(5y - 4x = 2\).

(a)
M1: For a correct method to find the midpoint, e.g., \(\left(\frac{-2 + 6}{2}, \frac{7 - 3}{2}\right)\) (allow one sign error).
A1: \((2, 2)\).

(b)
M1: For finding the gradient of \(PQ\) is \(-\frac{5}{4}\) (or \(-1.25\)) and stating the perpendicular gradient is \(\frac{4}{5}\) (or \(0.8\)).
M1: For substituting their midpoint and perpendicular gradient into \(y - y_1 = m(x - x_1)\) or \(y = mx + c\).
A1: \(5y - 4x = 2\) (or any non-zero integer multiple, e.g., \(10y - 8x = 4\)).

(a) Both chocolates of the same type means both are Milk, both are Dark, or both are White.
\(\text{P(Milk, Milk)} = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}\)
\(\text{P(Dark, Dark)} = \frac{4}{12} \times \frac{3}{11} = \frac{12}{132}\)
\(\text{P(White, White)} = \frac{3}{12} \times \frac{2}{11} = \frac{6}{132}\)
\(\text{Total Probability} = \frac{20 + 12 + 6}{132} = \frac{38}{132} = \frac{19}{66}\).

(b) Using the complement method, the probability of choosing no milk chocolates (only dark or white) from the 7 non-milk chocolates is:
\(\text{P(No Milk)} = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} = \frac{7}{22}\)
\(\text{P(At least one Milk)} = 1 - \frac{7}{22} = \frac{15}{22}\).

(a)
M1: For calculating the probability of any one same-type pair, e.g., \(\frac{5}{12} \times \frac{4}{11}\) or \(\frac{4}{12} \times \frac{3}{11}\) or \(\frac{3}{12} \times \frac{2}{11}\).
M1: For summing the three individual probabilities: \(\frac{20}{132} + \frac{12}{132} + \frac{6}{132}\).
A1: \(\frac{19}{66}\) (accept \(\frac{38}{132}\) or decimal \(0.288\) or \(0.29\) with working).

(b)
M1: For finding the probability of no milk chocolates: \(\frac{7}{12} \times \frac{6}{11}\) (\(=\frac{42}{132}\) or \(\frac{7}{22}\)), or for a complete direct sum of three correct cases: \(\left(\frac{5}{12} \times \frac{7}{11}\right) + \left(\frac{7}{12} \times \frac{5}{11}\right) + \left(\frac{5}{12} \times \frac{4}{11}\right)\).
A1: \(\frac{15}{22}\) (accept \(\frac{90}{132}\) or decimal \(0.682\) or \(0.68\) with working).

(a) Multiply both sides by \(3x - 4\):
\(w(3x - 4) = 2x + 5\)
\(3wx - 4w = 2x + 5\)
Rearrange to group all terms with \(x\) on one side:
\(3wx - 2x = 4w + 5\)
Factorise \(x\) on the left side:
\(x(3w - 2) = 4w + 5\)
Divide by \(3w - 2\):
\(x = \frac{4w + 5}{3w - 2}\).

(b) Write the left-hand side with a common denominator:
\(\frac{8}{2y} + \frac{3}{2y} = \frac{11}{8}\)
\(\frac{11}{2y} = \frac{11}{8}\)
By equating denominators:
\(2y = 8 \implies y = 4\).

(a)
M1: For multiplying by \((3x - 4)\) and expanding: \(3wx - 4w = 2x + 5\).
M1: For gathering terms in \(x\) on one side and factorising: \(x(3w - 2) = 4w + 5\).
A1: \(x = \frac{4w + 5}{3w - 2}\) (or equivalent, e.g., \(x = \frac{-4w - 5}{2 - 3w}\)).

(b)
M1: For combining the fractions to get \(\frac{11}{2y} = \frac{11}{8}\) or multiplying through by \(8y\) to get \(32 + 12 = 11y\).
A1: \(y = 4\).

(a) The area scale factor from \(A\) to \(B\) is:
\(\text{ASF} = \frac{405}{180} = 2.25\).
Thus, the linear scale factor (LSF) from \(A\) to \(B\) is:
\(\text{LSF} = \sqrt{2.25} = 1.5\).
Height of vase \(B\) = \(12 \times 1.5 = 18\text{ cm}\).

(b) The volume scale factor (VSF) from \(A\) to \(B\) is:
\(\text{VSF} = (1.5)^3 = 3.375\).
Volume of vase \(A\) = \(\frac{\text{Volume of } B}{\text{VSF}} = \frac{1350}{3.375} = 400\text{ cm}^3\).

(a)
M1: For finding the area scale factor \(\frac{405}{180} = 2.25\) (or \(\frac{180}{405} = \frac{4}{9}\)) and taking the square root to find the linear scale factor: \(\sqrt{2.25} = 1.5\) (or \(\sqrt{4/9} = 2/3\)).
A1: \(18\).

(b)
M1: For cubing their linear scale factor to find the volume scale factor: \((1.5)^3 = 3.375\) (or \((2/3)^3 = 8/27\)).
M1: For dividing the volume of \(B\) by their volume scale factor: \(\frac{1350}{3.375}\) (or \(1350 \times \frac{8}{27}\)).
A1: \(400\).

(a) First, find \(\text{g}(3)\):
\(\text{g}(3) = \frac{2}{3 + 1} = \frac{2}{4} = 0.5\).
Now substitute this value into \(\text{f}(x)\):
\(\text{fg}(3) = \text{f}(0.5) = 3(0.5) - 7 = 1.5 - 7 = -5.5\).

(b) Let \(y = \frac{2}{x + 1}\).
Rearrange to make \(x\) the subject of the formula:
\(y(x + 1) = 2\)
\(xy + y = 2\)
\(xy = 2 - y\)
\(x = \frac{2 - y}{y}\).
Thus, \(\text{g}^{-1}(x) = \frac{2 - x}{x}\).

(a)
M1: For finding \(\text{g}(3) = 0.5\) and substituting this into \(\text{f}(x)\), i.e., \(3(0.5) - 7\).
A1: \(-5.5\) (or \(-\frac{11}{2}\)).

(b)
M1: For setting \(y = \frac{2}{x + 1}\) (or \(x = \frac{2}{y + 1}\)) and multiplying by the denominator: \(y(x + 1) = 2\).
M1: For a correct rearrangement step to isolate \(x\) (or \(y\)), e.g., \(xy = 2 - y\).
A1: \(\text{g}^{-1}(x) = \frac{2 - x}{x}\) (or equivalent, e.g., \(\frac{2}{x} - 1\)).

We factorise each quadratic expression in the formula separately:

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

2. Factorise the first denominator:
\(2x^2 - 18 = 2(x^2 - 9) = 2(x - 3)(x + 3)\)

3. Factorise the second numerator:
\(2x^2 + 5x + 2 = (2x + 1)(x + 2)\)

4. Factorise the second denominator:
\(2x^2 + 14x + 24 = 2(x^2 + 7x + 12) = 2(x + 3)(x + 4)\)

Now, we rewrite the division as a multiplication by the reciprocal of the second fraction:

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

Next, we cancel out the common factors in the numerator and denominator:
- The term \((2x + 1)\) cancels out.
- The term \((x - 3)\) cancels out.
- The term \(2\) cancels out.
- The term \((x + 3)\) cancels out.

This leaves us with the simplified expression:

\(\frac{x + 4}{x + 2}\)

M1: for factorising \(2x^2 - 5x - 3\) to \((2x+1)(x-3)\)
M1: for factorising \(2x^2 - 18\) to \(2(x-3)(x+3)\) or \(2(x^2-9)\)
M1: for factorising \(2x^2 + 5x + 2\) to \((2x+1)(x+2)\)
M1: for factorising \(2x^2 + 14x + 24\) to \(2(x+3)(x+4)\) or \(2(x^2+7x+12)\)
M1: for inverting the second fraction and showing a multiplication with at least three correct factorisations
A1: for fully correct algebraic working leading to \(\frac{x+4}{x+2}\)

(a) The volume of a rectangular box with a square base of side length \(x\) and height \(h\) is given by:
\(V = x^2 h = 216\)

Rearranging for \(h\):
\(h = \frac{216}{x^2}\)

The total surface area \(A\) of a closed box is the sum of the areas of the 6 faces:
\(A = 2x^2 + 4xh\)

Substituting the expression for \(h\) into the surface area formula:
\(A = 2x^2 + 4x\left(\frac{216}{x^2}\right)\)

\(A = 2x^2 + \frac{864}{x}\) (as required).

(b) To find the minimum surface area, differentiate \(A\) with respect to \(x\):
\(\frac{dA}{dx} = 4x - 864x^{-2} = 4x - \frac{864}{x^2}\)

Set the derivative equal to 0 to find the stationary point:

\(4x - \frac{864}{x^2} = 0 \implies 4x = \frac{864}{x^2} \implies 4x^3 = 864 \implies x^3 = 216 \implies x = 6\)

To verify that this point is a minimum, find the second derivative:
\(\frac{d^2A}{dx^2} = 4 + 1728x^{-3} = 4 + \frac{1728}{x^3}\)

When \(x = 6\):
\(\frac{d^2A}{dx^2} = 4 + \frac{1728}{216} = 4 + 8 = 12\)

Since \(\frac{d^2A}{dx^2} > 0\), the stationary point is a minimum.

Calculate the minimum surface area by substituting \(x = 6\) back into the surface area equation:
\(A = 2(6)^2 + \frac{864}{6} = 72 + 144 = 216\text{ cm}^2\).

M1 (Part a): for writing a correct volume formula \(x^2 h = 216\) and making \(h\) the subject as \(h = \frac{216}{x^2}\)
A1 (Part a): for substituting \(h\) into the surface area formula \(A = 2x^2 + 4xh\) and simplifying to get the given result
M1 (Part b): for differentiating \(A\) correctly to get \(\frac{dA}{dx} = 4x - \frac{864}{x^2}\) (allow one minor differentiation slip)
M1 (Part b): for setting \(\frac{dA}{dx} = 0\) and solving to find \(x = 6\)
M1 (Part b): for finding the second derivative \(\frac{d^2A}{dx^2} = 4 + \frac{1728}{x^3}\) and substituting \(x = 6\) to show \(\frac{d^2A}{dx^2} > 0\)
A1 (Part b): for evaluating the minimum surface area to get \(216\)