2023 AQA GCSE Mathematics 8300 Practice Paper with Answers

First, rewrite the negative exponent as a reciprocal: \( 27^{-\frac{2}{3}} = \frac{1}{27^{2/3}} \). Next, apply the fractional exponent by taking the cube root and then squaring the result: \( 27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9 \). Therefore, the value is \( \frac{1}{9} \).

M1: Recognising that the negative power means reciprocal or the fractional power means cube root and squaring, e.g. showing \( \frac{1}{27^{2/3}} \) or \( \sqrt[3]{27} = 3 \). A1: Correct answer of \( \frac{1}{9} \).

Since \( y \) is inversely proportional to the square of \( x \), we can write the equation as \( y = \frac{k}{x^2} \). Use the given values to find the constant of proportionality, \( k \): \( 4 = \frac{k}{3^2} \implies 4 = \frac{k}{9} \implies k = 36 \). Now, use \( k = 36 \) to find \( y \) when \( x = 6 \): \( y = \frac{36}{6^2} = \frac{36}{36} = 1 \).

M1: Setting up the proportion equation \( y = \frac{k}{x^2} \) and finding \( k = 36 \). A1: Substituting \( x = 6 \) to get \( y = 1 \).

Let \( x = 0.2777... \). Multiplying by 10 gives \( 10x = 2.777... \). Multiplying by 100 gives \( 100x = 27.777... \). Subtracting these two equations: \( 100x - 10x = 27.777... - 2.777... \implies 90x = 25 \). Solving for \( x \) and simplifying the fraction: \( x = \frac{25}{90} = \frac{5}{18} \).

M1: Finding two equations that allow subtraction to eliminate the recurring part, e.g. \( 100x = 27.\dot{7} \) and \( 10x = 2.\dot{7} \), and subtracting them to get \( 90x = 25 \). A1: Correctly simplifying \( \frac{25}{90} \) to \( \frac{5}{18} \).

Factorise the numerator using the difference of two squares: \( x^2 - 9 = (x - 3)(x + 3) \). Factorise the denominator by taking out the common factor: \( 2x^2 + 6x = 2x(x + 3) \). Rewrite the fraction and cancel the common factor of \( (x + 3) \): \( \frac{(x - 3)(x + 3)}{2x(x + 3)} = \frac{x - 3}{2x} \).

M1: Factorising either the numerator to \( (x-3)(x+3) \) or the denominator to \( 2x(x+3) \). A1: Fully simplifying to \( \frac{x-3}{2x} \).

Subtract \( 3x \) from both sides: \( -5 > 4x + 11 \). Subtract \( 11 \) from both sides: \( -16 > 4x \). Divide both sides by \( 4 \): \( -4 > x \), which is equivalent to \( x < -4 \).

M1: Correctly rearranging terms to isolate the variable, e.g. obtaining \( -4x > 16 \) or \( -16 > 4x \). A1: Correct final inequality \( x < -4 \).

The probability of event A not happening is \( 1 - 0.3 = 0.7 \). The probability of event B not happening is \( 1 - 0.4 = 0.6 \). Since the events are independent, the probability of both not happening is the product of their individual probabilities: \( 0.7 \times 0.6 = 0.42 \).

M1: Calculating \( P(\text{not A}) = 0.7 \) and \( P(\text{not B}) = 0.6 \), and multiplying them. A1: Correct answer of 0.42.

The total surface area of a cylinder is given by the formula \( A = 2\pi r^2 + 2\pi r h \), where \( r \) is the radius and \( h \) is the height. Substitute the given values \( r = 3 \) and \( h = 10 \): Area of the two circular bases = \( 2 \times \pi \times 3^2 = 18\pi \). Curved surface area = \( 2 \times \pi \times 3 \times 10 = 60\pi \). Total surface area = \( 18\pi + 60\pi = 78\pi \text{ cm}^2 \).

M1: Correctly calculating either the area of the two circular bases (\( 18\pi \)) or the curved surface area (\( 60\pi \)). A1: Correct total surface area of \( 78\pi \text{ cm}^2 \).

First differences: \( 11-5=6 \), \( 19-11=8 \), \( 29-19=10 \). Second differences: \( 8-6=2 \), \( 10-8=2 \). Since the second difference is constant at 2, the coefficient of \( n^2 \) is \( 2/2 = 1 \). Subtract \( n^2 \) from each term of the sequence: For \( n=1 \): \( 5 - 1^2 = 4 \); For \( n=2 \): \( 11 - 2^2 = 7 \); For \( n=3 \): \( 19 - 3^2 = 10 \); For \( n=4 \): \( 29 - 4^2 = 13 \). The sequence of differences is 4, 7, 10, 13, ... which has a common difference of 3, so its \( n \)th term is \( 3n + 1 \). Combining these, the \( n \)th term of the sequence is \( n^2 + 3n + 1 \).

M1: Finding the constant second difference of 2, establishing that the leading term is \( n^2 \), and working out the linear difference sequence \( 4, 7, 10, 13 \). A1: Correctly obtaining the formula \( n^2 + 3n + 1 \).

Since \(y\) is directly proportional to the square of \(x\), we can write the equation:
\(y = k x^2\) for some constant \(k\).

Substitute the known values \(x = 3\) and \(y = 18\):
\(18 = k \times 3^2\)
\(18 = 9k\)
\(k = 2\)

Now use \(k = 2\) to find \(y\) when \(x = 5\):
\(y = 2 \times 5^2\)
\(y = 2 \times 25\)
\(y = 50\)

Therefore, the correct option is B.

B1: For the correct option B (50).

Let \(x = 0.2\dot{7} = 0.2777...\)

Multiply by 10:
\(10x = 2.777...\)

Multiply by 100:
\(100x = 27.777...\)

Subtract the first equation from the second:
\(100x - 10x = 27.777... - 2.777...\)
\(90x = 25\)

Solve for \(x\):
\(x = \frac{25}{90}\)

Simplify the fraction by dividing both the numerator and denominator by 5:
\(x = \frac{5}{18}\)

Therefore, the correct option is C.

B1: For the correct option C (\frac{5}{18}).

First, expand the numerator using the laws of indices:
\((4a^3)^2 = 4^2 \times (a^3)^2 = 16a^6\)

Next, divide by the denominator:
\(\frac{16a^6}{8a^4} = \frac{16}{8} \times a^{6-4} = 2a^2\)

Therefore, the correct option is B.

B1: For the correct option B (2a^2).

Since these are the only counters in the bag, the sum of all probabilities must equal 1:
\(P(\text{Red}) + P(\text{Blue}) + P(\text{Green}) = 1\)

Substitute the given values into the equation:
\(0.4 + 3x + x = 1\)
\(0.4 + 4x = 1\)
\(4x = 0.6\)
\(x = 0.15\)

Now, calculate the probability of picking a blue counter:
\(P(\text{Blue}) = 3x = 3 \times 0.15 = 0.45\)

Therefore, the correct option is C.

B1: For the correct option C (0.45).

First, express both ratios with a common value for the blue counters. The ratio of red to blue is \(2 : 3 = 8 : 12\). The ratio of blue to green is \(4 : 5 = 12 : 15\). Combining these, the ratio of red to blue to green is \(8 : 12 : 15\). The total number of parts is \(8 + 12 + 15 = 35\). Therefore, the fraction of blue counters is \(\frac{12}{35}\).

M1: For finding a combined ratio or scaling both ratios to have a common term for blue (e.g., \(8 : 12\) and \(12 : 15\)). A1: \(\frac{12}{35}\) or equivalent fraction.

First, address the negative exponent by taking the reciprocal of the fraction: \(\left( \frac{27}{8} \right)^{-\frac{2}{3}} = \left( \frac{8}{27} \right)^{\frac{2}{3}}\). Next, find the cube root: \(\sqrt[3]{\frac{8}{27}} = \frac{2}{3}\). Finally, square the result: \(\left( \frac{2}{3} \right)^2 = \frac{4}{9}\).

M1: For dealing with the negative index by taking the reciprocal, or correctly finding the cube root of the terms (e.g. seeing \(\frac{2}{3}\) or \(\frac{9}{4}\)). A1: \(\frac{4}{9}\) (or equivalent fraction / recurring decimal \(0.\dot{4}\)).

Let \(x = 0.4777...\) Multiply by 10: \(10x = 4.777...\) Multiply by 100: \(100x = 47.777...\) Subtract the two equations: \(100x - 10x = 47.777... - 4.777...\) which gives \(90x = 43\). Thus, \(x = \frac{43}{90}\). Since 43 is a prime number, the fraction cannot be simplified further.

M1: For setting up two appropriate recurring decimal equations and subtracting them to eliminate the recurring part, e.g. \(100x - 10x = 47.7... - 4.7...\). A1: \(\frac{43}{90}\).

To factorise fully, find the highest common factor of the terms \(12x^2 y\) and \(-18xy^2\). The highest common factor of 12 and 18 is 6. The highest common factor of \(x^2\) and \(x\) is \(x\). The highest common factor of \(y\) and \(y^2\) is \(y\). Taking \(6xy\) outside the bracket gives \(6xy(2x - 3y)\).

M1: For factorising out any common factor of at least degree 2, e.g., \(6x(2xy - 3y^2)\) or \(3xy(4x - 6y)\) or \(xy(12x - 18y)\). A1: \(6xy(2x - 3y)\).

Subtract 7 from both sides: \(-3x < 15\). Divide both sides by -3, remembering to reverse the inequality sign when dividing by a negative number: \(x > -5\).

M1: For subtracting 7 from both sides to get \(-3x < 15\) (or \(-3x = 15\) or \(3x > -15\)). A1: \(x > -5\) (accept \(-5 < x\)).

The interior angle and exterior angle of a regular polygon sum to \(180^\circ\). Therefore, the exterior angle is \(180^\circ - 162^\circ = 18^\circ\). The sum of the exterior angles of any polygon is \(360^\circ\). The number of sides is \(360^\circ \div 18^\circ = 20\).

M1: For finding the exterior angle \(18^\circ\) or setting up the equation \(\frac{(n-2) \times 180}{n} = 162\). A1: 20.

The total probability of all outcomes is 1. The probability of choosing blue or yellow is \(1 - 0.25 = 0.75\). Let the probability of choosing a yellow counter be \(y\). The probability of choosing a blue counter is \(4y\). Therefore, \(4y + y = 0.75\), which simplifies to \(5y = 0.75\), so \(y = 0.15\).

M1: For subtracting \(0.25\) from 1 to get \(0.75\) and setting up the ratio equation \(5y = 0.75\) (or equivalent division by 5). A1: \(0.15\) (accept \(\frac{3}{20}\) or \(15\%\)).

The formula for the area of a circle is \(\text{Area} = \pi r^2\). Given \(\pi r^2 = 64\pi\), we divide both sides by \(\pi\) to get \(r^2 = 64\), so the radius \(r = 8\text{ cm}\). The formula for the circumference is \(\text{Circumference} = 2\pi r\). Substituting \(r = 8\) gives \(\text{Circumference} = 2 \times \pi \times 8 = 16\pi\text{ cm}\).

M1: For establishing \(r^2 = 64\) and finding \(r = 8\). A1: \(16\pi\) (accept \(16 \times \pi\)).

Convert both mixed numbers to improper fractions:
\(2\frac{1}{3} = \frac{7}{3}\)
\(1\frac{2}{5} = \frac{7}{5}\)

Now divide the fractions:
\(\frac{7}{3} \div \frac{7}{5} = \frac{7}{3} \times \frac{5}{7} = \frac{5}{3}\)

Convert back to a mixed number:
\(\frac{5}{3} = 1\frac{2}{3}\)

M1: Convert both mixed numbers to improper fractions (\(\frac{7}{3}\) and \(\frac{7}{5}\)) OR show a correct multiplication method (e.g. \(\frac{7}{3} \times \frac{5}{7}\))

A1: Correct final answer of \(1\frac{2}{3}\) (do not accept \(\frac{5}{3}\) as the question specifies a mixed number)

Subtract \(3x\) from both sides of the inequality:
\(2x - 3 > 8\)

Add \(3\) to both sides:
\(2x > 11\)

Divide by \(2\):
\(x > 5.5\) (or \(x > \frac{11}{2}\))

M1: For a correct first step to collect like terms on one side, e.g. \(5x - 3x > 8 + 3\) or \(2x > 11\)

A1: For the correct final inequality: \(x > 5.5\) or \(x > \frac{11}{2}\)

The difference in ratio parts is:
\(7 - 3 = 4\) parts.

Since Ben receives \(\text{£}24\) more, these \(4\) parts are worth \(\text{£}24\).
Value of 1 part:
\(\text{£}24 \div 4 = \text{£}6\)

Total number of parts:
\(3 + 7 = 10\) parts.

Total money shared:
\(10 \times \text{£}6 = \text{£}60\)

M1: For finding the value of one part: \(24 \div (7 - 3) = 6\), or showing a correct scaling method (e.g. writing \(7x - 3x = 24\))

A1: Correct final total of 60 (or \(\text{£}60\))

The probability of getting Tails is \(1 - 0.3 = 0.7\).

To get exactly one Head in two spins, the outcomes can be Heads then Tails (HT) or Tails then Heads (TH):
\(P(\text{HT}) = 0.3 \times 0.7 = 0.21\)
\(P(\text{TH}) = 0.7 \times 0.3 = 0.21\)

Sum of these probabilities:
\(0.21 + 0.21 = 0.42\)

M1: For calculating the probability of one specific combination: \(0.3 \times 0.7 = 0.21\) (or equivalent shown in a tree diagram)

A1: For the correct final probability of \(0.42\)

We need to find two numbers that multiply to \(-18\) and add to \(-7\).
These numbers are \(-9\) and \(+2\).

Thus, the factorised expression is:
\((x - 9)(x + 2)\)

M1: For any expression of the form \((x \pm a)(x \pm b)\) where \(a \times b = -18\) or \(a + b = -7\), or for identifying the numbers \(-9\) and \(2\)

A1: Correct factorisation: \((x - 9)(x + 2)\) (or \((x + 2)(x - 9)\))

The formula for the area of a sector is:
\(\text{Area} = \frac{\theta}{360} \times \pi r^2\)

Substitute \(\theta = 120\) and \(r = 6\):
\(\text{Area} = \frac{120}{360} \times \pi \times 6^2\)
\(\text{Area} = \frac{1}{3} \times 36\pi = 12\pi\text{ cm}^2\)

M1: For substituting correct values into the sector area formula: \(\frac{120}{360} \times \pi \times 6^2\) (or equivalent fractional coefficient of \(\pi \times 36\))

A1: Correct final answer of \(12\pi\) (accept with units: \(12\pi\text{ cm}^2\))

Find the common difference between consecutive terms:
\(11 - 5 = 6\)
\(17 - 11 = 6\)

Since the difference is \(+6\), the expression contains \(6n\).

To find the constant term, find the term before the first term (the 'zero' term):
\(5 - 6 = -1\)

Therefore, the \(n\)-th term expression is:
\(6n - 1\)

M1: For finding a common difference of 6, indicated by \(6n\) in their final expression (e.g., \(6n + c\) where \(c\) is any constant)

A1: Correct final expression: \(6n - 1\)

First, find the cost of making 5 kg of the blend (using the ratio parts of 3 kg and 2 kg):
Cost of Assam tea = \(3 \text{ kg} \times \pounds 8/\text{kg} = \pounds 24\)
Cost of Earl Grey tea = \(2 \text{ kg} \times \pounds 13/\text{kg} = \pounds 26\)
Total cost for 5 kg of the blend = \(\pounds 24 + \pounds 26 = \pounds 50\)

Next, calculate the cost price per kg of the blend:
Cost price per kg = \(\frac{\pounds 50}{5 \text{ kg}} = \pounds 10/\text{kg}\)

Now, calculate the selling price per kg with a 35% profit:
Selling price per kg = \(\pounds 10 \times 1.35 = \pounds 13.50\)

Finally, calculate the selling price of a 400g (0.4 kg) bag:
Selling price of 400g = \(0.4 \text{ kg} \times \pounds 13.50/\text{kg} = \pounds 5.40\)

Total Marks: 3.8
- **M1**: For calculating the cost of a combined batch using the ratio, e.g., \(3 \times 8 + 2 \times 13 = 50\) (1.0 mark)
- **M1**: For finding the cost price per kg, e.g., \(50 \div 5 = 10\), or the cost price of 400g of the mixture, e.g., \(10 \times 0.4 = 4\) (1.0 mark)
- **M1**: For applying the 35% profit to either the price per kg or the price per 400g, e.g., \(10 \times 1.35 = 13.50\) or \(4 \times 1.35\) (1.0 mark)
- **A1**: For the correct final answer of \(\pounds 5.40\) (accept 5.40 or 5.4, but award 0.8 marks instead of 1.0 if units are completely wrong, though standard GCSE accepts 5.40) (0.8 marks)

We can simplify the expression inside the square root by factorising out \(15^2\):

\(15^2 \times 13^2 - 15^2 \times 5^2 = 15^2 \times (13^2 - 5^2)\)

Evaluate the subtraction inside the bracket:
\(13^2 - 5^2 = 169 - 25 = 144\)

Now substitute this back into the square root:
\(\sqrt{15^2 \times 144} = \sqrt{15^2} \times \sqrt{144} = 15 \times 12 = 180\)

Next, evaluate the denominator:
\(3^2 \times 2 = 9 \times 2 = 18\)

Finally, divide the numerator by the denominator:
\(\frac{180}{18} = 10\)

Total Marks: 3.8
- **M1**: For factorising or expanding the terms under the square root, e.g., recognizing \(15^2(13^2 - 5^2)\) or calculating \(38025 - 5625 = 32400\) (1.0 mark)
- **M1**: For correctly evaluating the square root, e.g., \(15 \times 12 = 180\) or \(\sqrt{32400} = 180\) (1.0 mark)
- **M1**: For evaluating the denominator as \(18\) (1.0 mark)
- **A1**: For the correct final answer of \(10\) (0.8 marks)

First, calculate the sale price of one Fiction book:
Fiction discount = \(15\%\) of \(\pounds 10 = \pounds 1.50\)
Fiction sale price = \(\pounds 10 - \pounds 1.50 = \pounds 8.50\)

Next, calculate the cost of 4 Fiction books:
Cost of 4 Fiction books = \(4 \times \pounds 8.50 = \pounds 34.00\)

Subtract this from the total spend to find the amount spent on Non-fiction books:
Spend on Non-fiction = \(\pounds 65.20 - \pounds 34.00 = \pounds 31.20\)

Now, calculate the sale price of one Non-fiction book:
Non-fiction discount = \(35\%\) of \(\pounds 16 = 0.35 \times 16 = \pounds 5.60\)
Non-fiction sale price = \(\pounds 16 - \pounds 5.60 = \pounds 10.40\)

Finally, calculate the number of Non-fiction books purchased:
Number of Non-fiction books = \(\frac{31.20}{10.40} = 3\)

Total Marks: 3.8
- **M1**: For finding the sale price of a Fiction book (\(\pounds 8.50\)) and the total cost for 4 of them (\(\pounds 34\)) (1.0 mark)
- **M1**: For subtracting the Fiction cost from the total spend to find \(\pounds 31.20\) (1.0 mark)
- **M1**: For finding the sale price of a Non-fiction book (\(\pounds 10.40\)) (1.0 mark)
- **A1**: For dividing \(31.20\) by \(10.40\) to get the correct integer answer of \(3\) (0.8 marks)

First, calculate the interior angle of a regular pentagon (\(a\)):
Sum of interior angles of a pentagon = \((5-2) \times 180^\circ = 540^\circ\)
Interior angle \(a = \frac{540^\circ}{5} = 108^\circ\)

Next, calculate the interior angle of a regular octagon (\(b\)):
Sum of interior angles of an octagon = \((8-2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ\)
Interior angle \(b = \frac{1080^\circ}{8} = 135^\circ\)

Now, calculate the interior angle of the regular \(n\)-sided polygon:
Interior angle = \(a + b - 103 = 108 + 135 - 103 = 140^\circ\)

To find \(n\), we can use the exterior angle of this polygon:
Exterior angle = \(180^\circ - 140^\circ = 40^\circ\)

Since the sum of exterior angles of any polygon is \(360^\circ\):
\(n = \frac{360^\circ}{40^\circ} = 9\)

Total Marks: 3.8
- **M1**: For finding the interior angle of a regular pentagon (\(108^\circ\)) or regular octagon (\(135^\circ\)) (1.0 mark)
- **M1**: For finding both interior angles (\(108^\circ\) and \(135^\circ\)) (1.0 mark)
- **M1**: For calculating the target interior angle \(140^\circ\) and finding the exterior angle \(40^\circ\), or setting up the equation \(\frac{(n-2) \times 180}{n} = 140\) (1.0 mark)
- **A1**: For the correct value of \(n = 9\) (0.8 marks)

First, calculate the probability of choosing a counter that is NOT red:
\(P(\text{Blue or Green}) = 1 - 0.4 = 0.6\)

The ratio of blue to green counters is \(5 : 7\), which means there are \(5 + 7 = 12\) parts in total for blue and green counters.

The fraction of these counters that are green is \(\frac{7}{12}\).

Now, calculate the probability of choosing a green counter:
\(P(\text{Green}) = \frac{7}{12} \times 0.6 = \frac{7}{12} \times \frac{6}{10} = \frac{42}{120} = 0.35\)

We are given that there are 21 green counters in the bag.
Let the total number of counters be \(T\).
\(0.35 \times T = 21\)
\(T = \frac{21}{0.35} = \frac{2100}{35} = 60\)

Total Marks: 3.8
- **M1**: For finding the combined probability of blue and green counters is \(0.6\) (1.0 mark)
- **M1**: For using the ratio to find the probability of choosing a green counter, e.g., \(\frac{7}{12} \times 0.6 = 0.35\) (or equivalent ratio fraction) (1.0 mark)
- **M1**: For setting up a correct equation/relation linking the number of green counters to the total, e.g., \(0.35 \times T = 21\) or showing that \(1 \text{ part} = 3\) counters in the blue/green subgroup (1.0 mark)
- **A1**: For the correct total of \(60\) (0.8 marks)

First, set up an equation for the area of the rectangle:
\(\text{Area} = \text{length} \times \text{width}\)
\(52 = (2x + 3)(x - 1)\)

Expand the right side:
\(52 = 2x^2 - 2x + 3x - 3\)
\(52 = 2x^2 + x - 3\)

Rearrange into a standard quadratic equation form:
\(2x^2 + x - 55 = 0\)

Factorise the quadratic equation:
We need two numbers that multiply to \(2 \times (-55) = -110\) and add to \(1\).
These numbers are \(11\) and \(-10\).
\(2x^2 - 10x + 11x - 55 = 0\)
\(2x(x - 5) + 11(x - 5) = 0\)
\((2x + 11)(x - 5) = 0\)

This gives solutions:
\(x = -5.5\) or \(x = 5\)

Since a physical measurement must be positive, \(x\) must be \(5\).

Now, calculate the length and width:
Length = \(2(5) + 3 = 13\text{ cm}\)
Width = \(5 - 1 = 4\text{ cm}\)

Finally, calculate the perimeter of the rectangle:
Perimeter = \(2 \times (\text{length} + \text{width}) = 2 \times (13 + 4) = 2 \times 17 = 34\text{ cm}\)

Total Marks: 3.8
- **M1**: For setting up the equation \((2x + 3)(x - 1) = 52\) and expanding to obtain \(2x^2 + x - 55 = 0\) (1.0 mark)
- **M1**: For solving the quadratic equation to find \(x = 5\) (1.0 mark)
- **M1**: For substituting \(x = 5\) back to find the dimensions of the rectangle as \(13\text{ cm}\) and \(4\text{ cm}\) (1.0 mark)
- **A1**: For the correct perimeter of \(34\) (0.8 marks)

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

\(\frac{3}{x-2} - \frac{2}{x+3} = \frac{3(x+3) - 2(x-2)}{(x-2)(x+3)}
\)

Expand the numerator:
\(3(x+3) - 2(x-2) = 3x + 9 - 2x + 4 = x + 13\)

Expand the denominator:
\((x-2)(x+3) = x^2 + 3x - 2x - 6 = x^2 + x - 6\)

So, the simplified fraction is:
\(\frac{x + 13}{x^2 + x - 6}\)

Comparing this to the form \(\frac{ax + b}{x^2 + cx + d}\), we get:
\(a = 1\) (since \(x = 1x\))
\(b = 13\)
\(c = 1\) (since \(x = 1x\))
\(d = -6\)

Now, calculate \(a + b + c + d\):
\(a + b + c + d = 1 + 13 + 1 + (-6) = 9\)

Total Marks: 3.8
- **M1**: For putting the terms over a common denominator, i.e., \(3(x+3) - 2(x-2)\) (1.0 mark)
- **M1**: For expanding and simplifying the numerator to \(x + 13\) (1.0 mark)
- **M1**: For expanding the denominator to \(x^2 + x - 6\) (1.0 mark)
- **A1**: For identifying \(a=1, b=13, c=1, d=-6\) and finding the correct sum \(9\) (0.8 marks)

First, find the total surface area of the cylinder:
\(\text{Surface Area of Cylinder} = 2\pi r^2 + 2\pi r h\)
With \(r = 4\):
\(\text{Surface Area} = 2\pi(4^2) + 2\pi(4)h = 32\pi + 8\pi h\)

Next, find the surface area of the sphere:
\(\text{Surface Area of Sphere} = 4\pi R^2\)
With \(R = 6\):
\(\text{Surface Area} = 4\pi(6^2) = 144\pi\)

Set these two surface areas equal to each other to solve for \(h\):
\(32\pi + 8\pi h = 144\pi\)
Divide the entire equation by \(\pi\):
\(32 + 8h = 144\)
\(8h = 112\)
\(h = 14\text{ cm}\)

Finally, calculate the volume of the cylinder:
\(\text{Volume of Cylinder} = \pi r^2 h\)
\(\text{Volume} = \pi \times 4^2 \times 14 = 16 \times 14 \times \pi = 224\pi\text{ cm}^3\)

Total Marks: 3.8
- **M1**: For writing down or using a correct expression for the total surface area of the cylinder in terms of \(\pi\) and \(h\), i.e., \(32\pi + 8\pi h\) (1.0 mark)
- **M1**: For calculating the surface area of the sphere as \(144\pi\) (1.0 mark)
- **M1**: For setting up the equation and correctly solving for \(h = 14\) (1.0 mark)
- **A1**: For the correct volume of \(224\pi\) (accept \(224\) if the question's requirement of the form \(k\pi\) is interpreted as writing the value of \(k\), but the standard is \(224\pi\)) (0.8 marks)

1. Find the total number of parts in the ratio:
\( 2 + 3 + 15 = 20 \) parts.

2. Find the volume of one part:
\( 60 \div 20 = 3 \) litres.

3. Calculate the volume needed for each component:
- Concentrated liquid: \( 2 \times 3 = 6 \) litres
- Vinegar: \( 3 \times 3 = 9 \) litres

4. Calculate the number of bottles required (rounding up to the nearest whole bottle where necessary):
- Concentrated liquid: \( 6 \div 1.5 = 4 \) bottles
- Vinegar: \( 9 \div 2 = 4.5 \) bottles, which rounds up to \( 5 \) bottles.

5. Calculate the total cost:
- Cost of concentrated liquid: \( 4 \times £4.50 = £18.00 \)
- Cost of vinegar: \( 5 \times £2.80 = £14.00 \)
- Total cost: \( £18.00 + £14.00 = £32.00 \).

M1: For finding the total parts (20) and dividing the total volume by 20, e.g., \( 60 \div 20 = 3 \).
M1: For calculating the required volumes of concentrated liquid (6 litres) and vinegar (9 litres).
M1: For correctly determining the number of whole bottles needed: 4 bottles of concentrated liquid and 5 bottles of vinegar (condone 4.5 if corrected to 5 later).
A1: Correct total cost of \( £32 \) (or 32, £32.00).

1. Set up expressions for the perimeters of both shapes:
- Perimeter of the rectangle: \( 2((3x + 5) + (x + 2)) = 2(4x + 7) = 8x + 14 \)
- Perimeter of the equilateral triangle: \( 3(2x + 6) = 6x + 18 \)

2. Equate the perimeters to solve for \( x \):
\( 8x + 14 = 6x + 18 \)
Subtract \( 6x \) from both sides:
\( 2x + 14 = 18 \)
Subtract 14 from both sides:
\( 2x = 4 \)
\( x = 2 \)

3. Substitute \( x = 2 \) back into the rectangle's dimensions:
- Length: \( 3(2) + 5 = 11 \text{ cm} \)
- Width: \( 2 + 2 = 4 \text{ cm} \)

4. Calculate the area of the rectangle:
\( \text{Area} = \text{length} \times \text{width} = 11 \times 4 = 44 \text{ cm}^2 \).

M1: For writing a correct expression for the perimeter of the rectangle, e.g., \( 8x + 14 \), or the perimeter of the triangle, e.g., \( 6x + 18 \).
M1: For setting up the equation \( 8x + 14 = 6x + 18 \) and solving to find \( x = 2 \).
M1: For substituting their value of \( x \) back into the expressions for length and width to find 11 and 4.
A1: Correct area of \( 44 \) (or \( 44 \text{ cm}^2 \)).

First, multiply the map distance by the scale factor to find the actual distance in centimetres:
\[ 8.4 \times 25\,000 = 210\,000\text{ cm} \]

Next, convert centimetres to metres by dividing by \(100\):
\[ 210\,000 \div 100 = 2100\text{ m} \]

Finally, convert metres to kilometres by dividing by \(1000\):
\[ 2100 \div 1000 = 2.1\text{ km} \]

This matches option B.

M1: For \(8.4 \times 25\,000\) or obtaining \(210\,000\) (or for dividing \(8.4\) by \(100\,000\) to convert map scale first).
A1: Correct answer of \(2.1\text{ km}\) (Option B).

The value after 3 years can be calculated using the compound depreciation formula:
\[ \text{Value} = 15\,000 \times (1 - 0.08)^3 \]
\[ \text{Value} = 15\,000 \times 0.92^3 \]
\[ \text{Value} = 15\,000 \times 0.778688 = 11\,680.32 \]

To the nearest pound, the value is \(\pounds 11\,680\), which corresponds to option B.

M1: For a correct compound interest/depreciation method, e.g., \(15\,000 \times 0.92^3\), or step-by-step calculations showing year-on-year reductions by \(8\%\).
A1: For rounding to \(\pounds 11\,680\) (Option B).

Find the highest common factor (HCF) of the two terms:
- The HCF of \(12\) and \(18\) is \(6\).
- The HCF of \(x^2\) and \(x\) is \(x\).
- The HCF of \(y\) and \(y^2\) is \(y\).

Thus, the HCF of the terms is \(6xy\).

Divide each term by \(6xy\) to find the terms inside the bracket:
\[ \frac{12x^2y}{6xy} = 2x \]
\[ \frac{18xy^2}{6xy} = 3y \]

So the fully factorised expression is:
\[ 6xy(2x - 3y) \]

This corresponds to option B.

M1: For finding a partial common factor (e.g., \(6\), \(xy\), or \(3xy\)) resulting in an expression like \(6(2x^2y - 3xy^2)\) or \(xy(12x - 18y)\).
A1: For fully factorising to \(6xy(2x - 3y)\) (Option B).

The sum of the probabilities of all outcomes must be \(1\).
\[ P(\text{blue or green}) = 1 - P(\text{red}) = 1 - 0.35 = 0.65 \]

The ratio of blue to green is \(2 : 3\), which means there are \(2 + 3 = 5\) equal parts in total for the remaining counters.

Green accounts for \(\frac{3}{5}\) of the combined probability of blue and green:
\[ P(\text{green}) = \frac{3}{5} \times 0.65 = 0.39 \]

This matches option B.

M1: For calculating the remaining probability \(1 - 0.35 = 0.65\) and attempting to divide by \(5\) or multiply by \(\frac{3}{5}\).
A1: Correct probability of \(0.39\) (Option B).

Rearrange the inequality to solve for \(x\):

Subtract \(3x\) from both sides:
\[ 7 < 2x - 3 \]

Add \(3\) to both sides:
\[ 10 < 2x \]

Divide both sides by \(2\):
\[ 5 < x \]

Which can also be written as:
\[ x > 5 \]

This matches option A.

M1: For a correct first step to collect like terms, e.g., subtracting \(3x\) from both sides to get \(2x\), or subtracting \(5x\) to get \(-2x\).
A1: Correctly solving to \(x > 5\) (Option A).

Let's find the differences between consecutive terms:
- First differences: \(5, 7, 9\)
- Second differences: \(2, 2\)

Since the second differences are constant and equal to \(2\), the coefficient of \(n^2\) is \(\frac{2}{2} = 1\). Thus, the formula starts with \(n^2\).

Now, subtract \(n^2\) from each term of the sequence:
- For \(n = 1\): \(3 - 1^2 = 2\)
- For \(n = 2\): \(8 - 2^2 = 4\)
- For \(n = 3\): \(15 - 3^2 = 6\)
- For \(n = 4\): \(24 - 4^2 = 8\)

The resulting linear sequence is \(2, 4, 6, 8, \dots\), which has the formula \(2n\).

Combining these gives the \(n\)-th term formula:
\[ n^2 + 2n \]

This matches option B.

M1: For finding the first differences (\(5, 7, 9\)) and identifying the second difference is \(2\), implying a term of \(n^2\), or for testing values of \(n\) in the given options.
A1: Correctly identifying \(n^2 + 2n\) (Option B).

The formula for the total surface area of a cylinder is:
\[ A = 2\pi r^2 + 2\pi r h \]

Substitute the given values (\(r = 4\), \(h = 10\)) into the formula:
\[ A = 2\pi (4)^2 + 2\pi (4)(10) \]
\[ A = 2\pi (16) + 80\pi \]
\[ A = 32\pi + 80\pi = 112\pi \]

Using a calculator:
\[ A \approx 351.858\text{ cm}^2 \]

Rounded to 3 significant figures, this is \(352\text{ cm}^2\).

This matches option C.

M1: For using the correct formula for total surface area, showing the sum of curved surface area and two circular face areas: \(2\pi(4)(10) + 2\pi(4^2)\).
A1: For the correct calculated surface area to 3 s.f. which is \(352\text{ cm}^2\) (Option C).

First, calculate the cumulative frequencies:
- For \(0 < t \le 5\): \(8\)
- For \(5 < t \le 10\): \(8 + 15 = 23\)
- For \(10 < t \le 15\): \(23 + 18 = 41\)
- For \(15 < t \le 20\): \(41 + 9 = 50\)

With \(50\) students, the median is located around the \(25\)-th or \(25.5\)-th position.

Since \(23\) students took \(10\) minutes or less, and \(41\) students took \(15\) minutes or less, the \(25\)-th and \(25.5\)-th values must lie in the interval \(10 < t \le 15\).

This matches option C.

M1: For calculating cumulative frequencies (\(8, 23, 41, 50\)) or identifying that the median position is at \(25\) or \(25.5\).
A1: Identifying the correct interval \(10 < t \le 15\) (Option C).

First, find the value of \(x\) using the fact that the sum of probabilities must equal 1:

\(0.35 + x + 0.2 + 2x = 1\)

\(0.55 + 3x = 1\)

\(3x = 0.45\)

\(x = 0.15\)

Now, calculate the probability of landing on 4:

\(P(4) = 2x = 2 \times 0.15 = 0.30\)

Finally, calculate the expected frequency for 400 spins:

\(\text{Expected frequency} = 0.30 \times 400 = 120\)

M1: Set up the probability equation \(0.35 + x + 0.2 + 2x = 1\) and solve for \(x = 0.15\) or find \(P(4) = 0.3\)
A1: Correct expected value of 120

First, find the distance traveled in each part of the journey:

Distance 1: \(50\text{ mph} \times 2\text{ hours} = 100\text{ miles}\)

Distance 2: \(70\text{ mph} \times 0.5\text{ hours} = 35\text{ miles}\)

Total distance = \(100 + 35 = 135\text{ miles}\)

Total time = \(2\text{ hours} + 0.5\text{ hours} = 2.5\text{ hours}\)

Average speed = \(\frac{\text{Total distance}}{\text{Total time}} = \frac{135\text{ miles}}{2.5\text{ hours}} = 54\text{ mph}\)

M1: Calculate total distance of 135 miles or identify total time of 2.5 hours
A1: Correct average speed of 54 mph

To find the amount of flour per muffin, divide the total flour by the number of muffins: 240 / 8 = 30 grams per muffin. Multiply this by 14 to find the amount needed for 14 muffins: 30 * 14 = 420 grams.

M1 for 240 / 8 or 30 or 14 / 8. A1 for 420.

The sale price represents 100% - 15% = 85% of the original price. Let x be the original price: 0.85 * x = 57.80. This gives x = 57.80 / 0.85 = 68.

M1 for 57.80 / 0.85 or 57.80 / 85 * 100. A1 for 68.

The radius of the circle is half the diameter: r = 12 / 2 = 6 cm. The formula for the area of a circle is A = pi * r^2. Substituting the values gives A = pi * 6^2 = 36 * pi which is approximately 113.097. To 3 significant figures, the area is 113.

M1 for pi * 6^2 or pi * (12/2)^2. A1 for 113.

Subtract 2x from both sides of the inequality to get 3x - 3 > 9. Then add 3 to both sides to get 3x > 12. Finally, divide both sides by 3 to get x > 4.

M1 for a correct first step to collect terms in x or constants, e.g. 3x - 3 > 9 or 5x - 2x > 9 + 3. A1 for x > 4.

The probability of choosing a green counter is 1 - (0.35 + 0.40) = 1 - 0.75 = 0.25. The number of green counters in the bag is 0.25 * 40 = 10.

M1 for 1 - (0.35 + 0.40) or 0.25 or finding the expected number of red (14) or blue (16) counters. A1 for 10.

The terms increase by 4 each time, so the expression starts with 4n. When n = 1, 4(1) = 4. To reach the first term 7, we must add 3. Therefore, the expression for the n-th term is 4n + 3.

M1 for finding a common difference of 4, seen in 4n or 4n + c. A1 for 4n + 3.

First, evaluate the expression inside the square root: 8.2^2 - 3.4^2 = 67.24 - 11.56 = 55.68. Then, calculate the square root of 55.68, which is approximately 7.4619. Rounded to 2 decimal places, this is 7.46.

M1 for 55.68 or showing intermediate evaluation of 8.2^2 - 3.4^2. A1 for 7.46.

10 cm is equal to 0.1 m. The accuracy limit is +/- half of the degree of accuracy, which is 0.1 / 2 = 0.05 m. The upper bound is 2.4 + 0.05 = 2.45 m.

M1 for identifying a half-width of 0.05 m or 5 cm, or attempting 2.4 + 0.05. A1 for 2.45.

First, find the total time in minutes for the initial production run:
\(2.5 \text{ hours} = 2.5 \times 60 = 150 \text{ minutes}\).

Now, calculate the rate of production in minutes per component:
\(\text{Time per component} = \frac{150 \text{ minutes}}{350 \text{ components}} = \frac{3}{7} \text{ minutes per component}\).

Finally, calculate the time required for \(98\) components:
\(\text{Time} = 98 \times \frac{3}{7} = 14 \times 3 = 42 \text{ minutes}\).

M1: For a correct method to find the rate, e.g., \(2.5 \times 60 = 150\) or \(150 \div 350\) (or \(350 \div 150 = 2.33...\) components per minute)
A1: 42

Let the original price of the coat be \(x\).
An reduction of \(15\%\) means the sale price is \(85\%\) of the original price.
\(0.85x = 57.80\)
\(x = \frac{57.80}{0.85} = 68\).
The original price of the coat was \(£68\).

M1: For \(57.80 \div 0.85\) or \(\frac{57.80}{85} \times 100\) (or equivalent)
A1: 68 (or £68)

Multiply both sides of the equation by \(4\) to clear the fraction:
\(2x - 3 = 4(x - 6)\)

Expand the bracket on the right side:
\(2x - 3 = 4x - 24\)

Subtract \(2x\) from both sides:
\(-3 = 2x - 24\)

Add \(24\) to both sides:
\(21 = 2x\)

Divide by \(2\):
\(x = 10.5\) (or \(\frac{21}{2}\))

M1: For a correct step to eliminate the fraction, e.g., \(2x - 3 = 4(x - 6)\) or \(2x - 3 = 4x - 24\)
A1: 10.5 (or equivalent fraction, e.g., 21/2)

First, determine the upper bounds for the length and the width:
- For the length (to the nearest \(1\text{ m}\)), the upper bound is \(85 + 0.5 = 85.5\text{ m}\).
- For the width (to the nearest \(5\text{ m}\)), the upper bound is \(40 + 2.5 = 42.5\text{ m}\).

Calculate the upper bound for the area by multiplying these upper bounds together:
\(\text{Upper Bound of Area} = 85.5 \times 42.5 = 3633.75\text{ m}^2\).

M1: For identifying \(85.5\) or \(42.5\) as an upper bound
A1: 3633.75

First, find the relative frequency of landing on \(4\). Since the total relative frequency must equal \(1\):
\(P(4) = 1 - (0.24 + 0.35 + 0.18) = 1 - 0.77 = 0.23\)

Now, calculate the expected number of times it lands on \(4\) in \(200\) spins:
\(\text{Expected Frequency} = 200 \times 0.23 = 46\).

M1: For \(1 - (0.24 + 0.35 + 0.18)\) or \(0.23\) seen
A1: 46

Expand the double brackets by multiplying each term in the first bracket by each term in the second bracket:
\((3x - 4)(2x + 5) = (3x \times 2x) + (3x \times 5) + (-4 \times 2x) + (-4 \times 5)\)
\(= 6x^2 + 15x - 8x - 20\)

Combine the like terms (the \(x\) terms):
\(= 6x^2 + 7x - 20\)

M1: For at least 3 correct terms out of 4 in the expansion (e.g., \(6x^2, 15x, -8x, -20\))
A1: \(6x^2 + 7x - 20\)

Let the mass of copper in a sample of the alloy be \(5x\text{ g}\) and the mass of tin be \(3x\text{ g}\).
Total mass of the alloy = \(5x + 3x = 8x\text{ g}\).
Using the formula \(\text{Volume} = \frac{\text{Mass}}{\text{Density}}\):
Volume of copper = \(\frac{5x}{8.9}\text{ cm}^3\)
Volume of tin = \(\frac{3x}{7.3}\text{ cm}^3\)
Total volume of the alloy = \(\frac{5x}{8.9} + \frac{3x}{7.3} \approx 0.5618x + 0.4110x = 0.9728x\text{ cm}^3\).
Density of the alloy = \(\frac{\text{Total Mass}}{\text{Total Volume}} = \frac{8x}{\frac{5x}{8.9} + \frac{3x}{7.3}} = \frac{8}{\frac{5}{8.9} + \frac{3}{7.3}} \approx 8.224\text{ g/cm}^3\).
To 2 decimal places, the density is \(8.22\text{ g/cm}^3\).

M1 for expressing volume of copper as \(\frac{5x}{8.9}\) or volume of tin as \(\frac{3x}{7.3}\) (or using specific values like 50g and 30g)
M1 for total volume \(\approx 0.973x\) or \(\frac{5}{8.9} + \frac{3}{7.3}\)
M1 for total mass divided by total volume \(\frac{8}{\text{total volume}}\)
A1 for \(8.22\text{ g/cm}^3\) (accept answers in the range 8.22 to 8.23)

Let the original price of the jacket be \(P\).
After the first reduction of \(15\%\), the price is \(P \times (1 - 0.15) = 0.85P\).
In the final week, this price is reduced by \(\frac{1}{10}\), which is a \(10\%\) reduction.
The final price is \(0.85P \times \left(1 - \frac{1}{10}\right) = 0.85P \times 0.9 = 0.765P\).
We are given that this final price is £61.20:
\(0.765P = 61.20\)
\(P = \frac{61.20}{0.765} = 80\).
So, the original price was £80.

M1 for representing the price after the first reduction as \(0.85P\) or equivalent
M1 for representing the final price as \(0.85P \times 0.9\) or \(0.765P\) or setting up the equation \(0.9 \times 0.85P = 61.20\)
M1 for a complete method to solve for \(P\), i.e., \(\frac{61.20}{0.765}\)
A1 for \(80\) or £80

The area of a rectangle is \(\text{length} \times \text{width}\).
Set up the equation for area:
\((2x + 3)(x - 1) = 33\)
Expand the brackets:
\(2x^2 - 2x + 3x - 3 = 33\)
\(2x^2 + x - 3 = 33\)
Rearrange to form a quadratic equation equal to 0:
\(2x^2 + x - 36 = 0\)
Factorise the quadratic expression:
\((2x + 9)(x - 4) = 0\)
This gives two possible solutions for \(x\):
\(x = -4.5\) or \(x = 4\).
Since dimensions must be positive, \(x = 4\).
Now find the length and width:
Length = \(2(4) + 3 = 11\text{ cm}\)
Width = \(4 - 1 = 3\text{ cm}\).
Perimeter = \(2 \times (\text{length} + \text{width}) = 2 \times (11 + 3) = 28\text{ cm}\).

M1 for setting up the equation \((2x + 3)(x - 1) = 33\) and expanding to \(2x^2 + x - 3 = 33\)
M1 for rearranging to \(2x^2 + x - 36 = 0\) and attempting to solve (by factorising or using the quadratic formula)
M1 for selecting \(x = 4\) and calculating the dimensions (length = 11, width = 3)
A1 for \(28\text{ cm}\)

The volume of a cylinder is \(V_{\text{cylinder}} = \pi r^2 h\).
Given \(h = 2r\), we have:
\(V_{\text{cylinder}} = \pi r^2 (2r) = 2\pi r^3\).
The volume of a sphere is \(V_{\text{sphere}} = \frac{4}{3}\pi r^3\).
The difference between their volumes is:
\(V_{\text{cylinder}} - V_{\text{sphere}} = 2\pi r^3 - \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3\).
We are given this difference is \(150\text{ cm}^3\):
\(\frac{2}{3}\pi r^3 = 150\)
Multiply both sides by 3 and divide by 2:
\(\pi r^3 = 225\)
\(r^3 = \frac{225}{\pi}\)
\(r = \sqrt[3]{\frac{225}{\pi}} \approx \sqrt[3]{71.6197} \approx 4.1528\text{ cm}\).
To 3 significant figures, \(r = 4.15\text{ cm}\).

M1 for writing the cylinder volume as \(2\pi r^3\) and sphere volume as \(\frac{4}{3}\pi r^3\)
M1 for setting up the difference equation \(\frac{2}{3}\pi r^3 = 150\) or equivalent
M1 for rearranging to find \(r = \sqrt[3]{\frac{225}{\pi}}\)
A1 for \(4.15\) (accept answers in the range 4.15 to 4.16)

Using the formula for the \(n\)-th term:
For \(n = 2\):
\(u_2 = a(2)^2 + b(2) + 5 = 15 \Rightarrow 4a + 2b + 5 = 15 \Rightarrow 4a + 2b = 10 \Rightarrow 2a + b = 5\) (Equation 1).
For \(n = 5\):
\(u_5 = a(5)^2 + b(5) + 5 = 60 \Rightarrow 25a + 5b + 5 = 60 \Rightarrow 25a + 5b = 55 \Rightarrow 5a + b = 11\) (Equation 2).
Subtract Equation 1 from Equation 2:
\((5a + b) - (2a + b) = 11 - 5\)
\(3a = 6 \Rightarrow a = 2\).
Substitute \(a = 2\) into Equation 1:
\(2(2) + b = 5 \Rightarrow 4 + b = 5 \Rightarrow b = 1\).
So, the \(n\)-th term of the sequence is \(u_n = 2n^2 + n + 5\).
To find the 10th term, substitute \(n = 10\):
\(u_{10} = 2(10)^2 + 10 + 5 = 2(100) + 10 + 5 = 200 + 10 + 5 = 215\).

M1 for substituting \(n=2\) and \(n=5\) to get two equations in terms of \(a\) and \(b\) (e.g. \(4a+2b=10\) and \(25a+5b=55\))
M1 for a valid method to solve the simultaneous equations to find \(a\) and \(b\)
M1 for finding \(a = 2\) and \(b = 1\)
A1 for \(215\)

Since there are only red, blue, and green counters, the sum of their probabilities must be 1:
\(P(\text{Blue}) = 1 - P(\text{Red}) - P(\text{Green}) = 1 - 0.35 - 0.25 = 0.40\).
Let \(N\) be the total number of counters in the bag.
Number of blue counters = \(0.40N\).
Number of green counters = \(0.25N\).
We are told there are 9 more blue counters than green counters:
\(0.40N - 0.25N = 9\)
\(0.15N = 9\)
\(N = \frac{9}{0.15} = 60\).
So, there are 60 counters in total.
The number of red counters is:
\(0.35 \times 60 = 21\).

M1 for finding the probability of picking a blue counter: \(1 - 0.35 - 0.25 = 0.40\)
M1 for setting up an equation for the difference in counters: \(0.40N - 0.25N = 9\) (or equivalent)
M1 for solving to find the total number of counters \(N = 60\)
A1 for \(21\)

First, find the initial total height of the girls:
\(\text{Total height of girls} = 15 \times 162 = 2430\text{ cm}\).
After two girls leave, the new total height of the girls is:
\(2430 - 155 - 168 = 2107\text{ cm}\).
The number of girls remaining is \(15 - 2 = 13\).

Next, find the initial total height of the boys:
\(\text{Total height of boys} = 10 \times 175 = 1750\text{ cm}\).
Three new boys join with a mean height of \(180\text{ cm}\), so their total height is:
\(3 \times 180 = 540\text{ cm}\).
The new total height of the boys is:
\(1750 + 540 = 2290\text{ cm}\).
The number of boys now is \(10 + 3 = 13\).

Now, calculate the combined total height of all children:
\(\text{Combined total height} = 2107 + 2290 = 4397\text{ cm}\).
The total number of children is \(13\text{ girls} + 13\text{ boys} = 26\text{ children}\).
Calculate the new overall mean height:
\(\text{New mean height} = \frac{4397}{26} \approx 169.115\text{ cm}\).
To 1 decimal place, the new mean height is \(169.1\text{ cm}\).

M1 for finding the initial total height of girls (2430) and boys (1750)
M1 for adjusting the totals: new girls' total = 2107 and new boys' total = 2290
M1 for dividing the combined total height (4397) by the total number of children (26)
A1 for \(169.1\)

To find the upper bound for the average speed, we use the formula:
\(\text{Speed}_{\text{UB}} = \frac{\text{Distance}_{\text{UB}}}{\text{Time}_{\text{LB}}}\)
First, find the upper bound for the distance \(d\):
Since \(d = 140\text{ miles}\) is correct to the nearest 5 miles, the upper bound is:
\(140 + 2.5 = 142.5\text{ miles}\)
Next, find the lower bound for the time \(t\):
Since \(t = 2.5\text{ hours}\) is correct to the nearest 0.1 hours, the lower bound is:
\(2.5 - 0.05 = 2.45\text{ hours}\)
Now, calculate the upper bound for the average speed:
\(\text{Speed}_{\text{UB}} = \frac{142.5}{2.45} \approx 58.163\text{ mph}\).
To 1 decimal place, the upper bound for the average speed is \(58.2\text{ mph}\).

M1 for finding the upper bound of distance as \(142.5\)
M1 for finding the lower bound of time as \(2.45\)
M1 for dividing their upper bound of distance by their lower bound of time
A1 for \(58.2\)

1. Find the required mass of each ingredient in the 1200 kg mixture: Total parts = 2 + 3 + 5 = 10. Mass of cement = \( \frac{2}{10} \times 1200 = 240 \) kg. Mass of sand = \( \frac{3}{10} \times 1200 = 360 \) kg. Mass of gravel = \( \frac{5}{10} \times 1200 = 600 \) kg. 2. Calculate the original cost of each material: Cost of cement = \( 240 \times 4.20 = 1008 \) pounds, Cost of sand = \( 360 \times 1.80 = 648 \) pounds, Cost of gravel = \( 600 \times 1.20 = 720 \) pounds. 3. Apply the 15% discount to the combined cost of cement and gravel: Combined cost of cement and gravel = \( 1008 + 720 = 1728 \) pounds. Discounted cost = \( 1728 \times 0.85 = 1468.80 \) pounds. 4. Calculate the total cost including sand and the delivery fee: Total cost = \( 1468.80 + 648 + 45 = 2161.80 \) pounds.

M1: Method to find the mass of at least one ingredient (e.g. \( \frac{2}{10} \times 1200 \) or 240 or 360 or 600). M1: Method to calculate the cost of cement and gravel with 15% discount (e.g. \( (1008 + 720) \times 0.85 \)). M1: Complete method to find total cost by adding sand cost and delivery fee (e.g. \( 1468.80 + 648 + 45 \)). A1: Correct total cost of 2161.80.

1. Find the volume of the prism from its mass and density: Volume = Mass / Density = \( 9676.8 / 8.96 = 1080 \) cm\(^3\). 2. Set up an equation for the volume of the prism: Area of cross-section = \( \frac{x + (x + 4)}{2} \times 6 = (2x + 4) \times 3 = 6x + 12 \) cm\(^2\). Volume = Area \( \times \) length = \( (6x + 12) \times 15 = 90x + 180 \) cm\(^3\). 3. Equate the volume and solve for \( x \): \( 90x + 180 = 1080 \), which gives \( 90x = 900 \), so \( x = 10 \).

M1: Method to calculate the volume using mass/density (e.g. \( 9676.8 / 8.96 = 1080 \)). M1: Method to write an algebraic expression for the area of the trapezium or the volume of the prism (e.g. \( 6x + 12 \) or \( 90x + 180 \)). M1: Setting up and simplifying a linear equation of the form \( ax + b = 1080 \). A1: Correct value of \( x = 10 \).

1. Find the probability of choosing a blue or yellow counter: \( P(\text{Blue or Yellow}) = 1 - 0.3 = 0.7 \). 2. Use the ratio of blue to yellow (4 : 3) to find the individual probability of choosing a blue counter: \( P(\text{Blue}) = 0.7 \times \frac{4}{4 + 3} = 0.7 \times \frac{4}{7} = 0.4 \). 3. Use the actual number of blue counters to find the total number of counters in the bag: Let \( N \) be the total number of counters. \( 0.4 \times N = 32 \implies N = 80 \). 4. Calculate the number of red counters: Number of red counters = \( 0.3 \times 80 = 24 \).

M1: Find the combined probability of blue and yellow (e.g. \( 1 - 0.3 = 0.7 \)). M1: Method to find the probability of choosing a blue counter (e.g. \( 0.7 \times \frac{4}{7} = 0.4 \)). M1: Method to find the total number of counters (e.g. \( 32 \div 0.4 = 80 \)). A1: Correct answer of 24.

The multiplier for a 12% annual depreciation is \(1 - 0.12 = 0.88\). Let \(P\) be the original value of the car. After 3 years, the value is given by \(P \times 0.88^3 = 11943.04\). Calculating the power: \(0.88^3 = 0.681472\). Now, divide to find \(P\): \(P = \frac{11943.04}{0.681472} = 17500\). Thus, the original value was £17,500.

M1: For setting up the equation \(P \times 0.88^3 = 11943.04\) or finding \(0.88^3 = 0.681472\). A0.1: For the correct final answer of £17,500.

Multiply both sides by \(2x + 1\): \(y(2x + 1) = 3x - 5\). Expand the left side: \(2xy + y = 3x - 5\). Rearrange to bring all terms containing \(x\) to one side: \(2xy - 3x = -y - 5\). Factorise \(x\) out: \(x(2y - 3) = -(y + 5)\). Divide both sides by \(2y - 3\): \(x = \frac{-(y + 5)}{2y - 3}\). Multiplying the numerator and denominator by \(-1\) gives \(x = \frac{y + 5}{3 - 2y}\).

M1: For expanding \(y(2x+1)\) and collecting terms with \(x\) on one side. A0.1: For completing the steps to obtain \(x = \frac{y + 5}{3 - 2y}\).

First, calculate the volume of the cylinder: \(V = \pi r^2 h = \pi \times 4^2 \times 9 = 144\pi\text{ cm}^3\). The volume of a sphere is given by \(V = \frac{4}{3}\pi R^3\). Since the volumes are equal: \(\frac{4}{3}\pi R^3 = 144\pi\). Divide both sides by \(\pi\): \(\frac{4}{3}R^3 = 144\). Multiply by \(\frac{3}{4}\): \(R^3 = 108\). Find the cube root of 108: \(R = \sqrt[3]{108} \approx 4.7622\text{ cm}\). To 3 significant figures, this is 4.76 cm.

M1: For setting up the volume equality \(\frac{4}{3}\pi R^3 = 144\pi\) and solving for \(R^3 = 108\). A0.1: For the correct radius of 4.76 cm.

Let \(H\) be the set of History students and \(G\) be the set of Geography students. The number of students studying at least one of these subjects is \(30 - 5 = 25\). Using the formula for union: \(n(H \cup G) = n(H) + n(G) - n(H \cap G)\), we get \(25 = 18 + 15 - n(H \cap G)\), which gives \(n(H \cap G) = 8\). The probability that a student studies Geography given they study History is \(P(G|H) = \frac{n(H \cap G)}{n(H)} = \frac{8}{18} = \frac{4}{9}\).

M1: For calculating the number of students studying both subjects as 8. A0.1: For expressing the conditional probability as \(\frac{4}{9}\).

First, find the critical values by factorising the quadratic expression \(2x^2 - 5x - 3 = 0\). This factorises to \((2x + 1)(x - 3) = 0\), giving critical values of \(x = -0.5\) and \(x = 3\). Since the inequality is \(> 0\), the solution is the region outside the critical values: \(x < -0.5\) or \(x > 3\).

M1: For factorising to \((2x+1)(x-3)\) and finding the critical values \(-0.5\) and \(3\). A0.1: For selecting the correct outside region \(x < -0.5\) or \(x > 3\).

Find the first differences: \(12-5=7\), \(23-12=11\), \(38-23=15\), \(57-38=19\). The second differences are constant at \(4\). Therefore, the coefficient of \(n^2\) is \(4 / 2 = 2\). Subtract \(2n^2\) from each term in the sequence: for \(n=1\), \(5 - 2(1) = 3\); for \(n=2\), \(12 - 2(4) = 4\); for \(n=3\), \(23 - 2(9) = 5\); for \(n=4\), \(38 - 2(16) = 6\). The resulting sequence is \(3, 4, 5, 6, ...\), which is a linear sequence with \(n\)-th term \(n + 2\). Combining these gives the final formula: \(2n^2 + n + 2\).

M1: For finding the second difference is 4, indicating a \(2n^2\) term. A0.1: For completing the process to find the correct \(n\)-th term expression \(2n^2 + n + 2\).

To find the turning point, complete the square on the quadratic expression: \(y = x^2 - 8x + 11 = (x - 4)^2 - 4^2 + 11 = (x - 4)^2 - 16 + 11 = (x - 4)^2 - 5\). The equation in the form \(y = (x - a)^2 + b\) has its turning point at \((a, b)\). Thus, the turning point is \((4, -5)\).

M1: For completing the square correctly to get \((x-4)^2-5\). A0.1: For stating the coordinates of the turning point as \((4, -5)\).

Find the midpoints of the weight intervals: 5, 15, 25, and 35. Multiply each midpoint by its corresponding frequency: \(4 \times 5 = 20\), \(8 \times 15 = 120\), \(5 \times 25 = 125\), \(3 \times 35 = 105\). Sum these products: \(20 + 120 + 125 + 105 = 370\). Divide this sum by the total frequency: \(\frac{370}{20} = 18.5\).

M1: For calculating the sum of products using midpoints to get 370. A0.1: For dividing by 20 to get the correct estimate of 18.5.

Step 1: Convert £350 to Euros. \(350 \times 1.16 = €406\). Step 2: Subtract the amount spent. \(406 - 220 = €186\). Step 3: Convert the remaining €186 back to Pounds. \(186 \times 0.82 = £152.52\).

M1 for calculating initial Euros (406) or finding remaining Euros (186). A1 for £152.52 (accept 152.52).

Step 1: Calculate the value after the first year. \(15000 \times 0.88 = £13,200\). Step 2: Calculate the value after the second year. \(13200 \times 0.92 = £12,144\).

M1 for multiplying by 0.88 or 0.92, or finding 13200. A1 for £12,144 (accept 12144).

The perimeter of a rectangle is given by \(2(\text{length} + \text{width})\). So, \(2((3x + 2) + (x + 5)) = 46\). This simplifies to \(2(4x + 7) = 46\), which expands to \(8x + 14 = 46\). Subtracting 14 from both sides gives \(8x = 32\). Dividing by 8 gives \(x = 4\).

M1 for setting up a correct equation, e.g., \(2(4x + 7) = 46\) or \(8x + 14 = 46\). A1 for \(x = 4\).

The total probability of choosing any counter is 1. The probability of choosing either a red or blue counter is \(1 - 0.25 = 0.75\). Since the ratio of red to blue is \(3 : 2\), the fraction of the remaining probability representing blue is \(\frac{2}{3 + 2} = \frac{2}{5}\). Thus, the probability of choosing a blue counter is \(\frac{2}{5} \times 0.75 = 0.3\).

M1 for finding the remaining probability of 0.75 or writing \(\frac{2}{5} \times 0.75\). A1 for 0.3 (accept 3/10 or 30%).

The area of a sector is given by the formula \(\frac{\theta}{360} \times \pi r^2\). Substituting the given values: \(\text{Area} = \frac{72}{360} \times \pi \times 8.5^2 = 0.2 \times 72.25 \times \pi = 14.45\pi \approx 45.396\text{ cm}^2\). Rounded to 3 significant figures, this is 45.4.

M1 for a correct substitution into the sector area formula, e.g., \(\frac{72}{360} \times \pi \times 8.5^2\). A1 for 45.4 (accept answers in the range 45.39 to 45.41).

Step 1: Find the total number of goals scored: \((0 \times 3) + (1 \times 6) + (2 \times 7) + (3 \times 4) = 0 + 6 + 14 + 12 = 32\) goals. Step 2: Divide the total goals by the total number of matches (20): \(\frac{32}{20} = 1.6\).

M1 for finding the total number of goals (32) or showing a correct method to find the sum of products. A1 for 1.6.

The sequence increases by 6 each time, so the common difference is 6. The \(n\)-th term formula is \(a + (n - 1)d\). Here, \(a = 7\) and \(d = 6\), so the \(n\)-th term is \(7 + 6(n - 1) = 6n + 1\). To find the 100th term, substitute \(n = 100\): \(6(100) + 1 = 601\).

M1 for finding the \(n\)-th term expression \(6n + 1\) or for writing \(7 + 99 \times 6\). A1 for 601.

Since the measurements are correct to the nearest 5 m, the degree of accuracy is 5 m. The upper and lower bounds are found by adding or subtracting half of this unit (2.5 m). The lower bound of the length is \(80 - 2.5 = 77.5\) m. The lower bound of the width is \(50 - 2.5 = 47.5\) m. The lower bound of the area is \(77.5 \times 47.5 = 3681.25\) \(\text{m}^2\).

M1 for identifying the correct lower bounds for length (77.5) and width (47.5). A1 for 3681.25.

France price in pounds = €120 / 1.18 = £101.6949... Difference = £101.6949... - £95 = £6.6949... To 2 decimal places, this is £6.69.

M1 for 120 / 1.18 or 95 * 1.18 A1 for 6.69

The sale price represents 100% - 15% = 85% of the original price. Original price = £578 / 0.85 = £680.

M1 for 578 / 0.85 or equating 85% to 578 A1 for 680

Expand the first bracket: \(3(2x - 5) = 6x - 15\). Expand the second bracket: \(-2(x - 4) = -2x + 8\). Combine like terms: \(6x - 2x - 15 + 8 = 4x - 7\).

M1 for expanding at least one bracket correctly to get \(6x - 15\) or \(-2x + 8\) A1 for \(4x - 7\)

The probability of not landing on a 6 is \(1 - 0.28 = 0.72\). The expected number of times is \(0.72 \times 300 = 216\).

M1 for \(1 - 0.28\) or \(0.28 \times 300\) A1 for 216

The common difference is \(11 - 5 = 6\), so the sequence is related to \(6n\). When \(n = 1\), \(6(1) = 6\). To get 5, we subtract 1. Thus, the \(n\)th term is \(6n - 1\).

M1 for \(6n + c\) where \(c\) is any integer, or for identifying a common difference of 6 A1 for \(6n - 1\)

The lowest value that rounds to 7.4 to 1 decimal place is 7.35. The upper bound is 7.45. Therefore, the error interval is \(7.35 \le x < 7.45\).

M1 for identifying 7.35 or 7.45 as the bounds A1 for \(7.35 \le x < 7.45\)

Multiply both sides by 2: \(3x + 1 = 16\). Subtract 1 from both sides: \(3x = 15\). Divide by 3: \(x = 5\).

M1 for \(3x + 1 = 16\) A1 for 5

Volume of a prism = area of cross-section \(\times\) length. Volume = \(18 \times 7.5 = 135\text{ cm}^3\).

M1 for \(18 \times 7.5\) A1 for 135

First, calculate the mass of copper and zinc in \(400\text{ g}\) of Brass. Total parts = \(11 + 9 = 20\). Mass of copper in Brass = \(\frac{11}{20} \times 400 = 220\text{ g}\). Mass of zinc in Brass = \(\frac{9}{20} \times 400 = 180\text{ g}\). Next, calculate the mass of copper and zinc in \(600\text{ g}\) of Alloy \(B\). Total parts = \(7 + 3 = 10\). Mass of copper in Alloy \(B\) = \(\frac{7}{10} \times 600 = 420\text{ g}\). Mass of zinc in Alloy \(B\) = \(\frac{3}{10} \times 600 = 180\text{ g}\). Now, find the total mass of each metal in Alloy \(C\). Total copper = \(220 + 420 = 640\text{ g}\). Total zinc = \(180 + 180 = 360\text{ g}\). Write as a ratio and simplify: \(640 : 360 = 64 : 36 = 16 : 9\).

1 mark: Method to find mass of copper or zinc in Brass (e.g., \(220\text{ g}\) or \(180\text{ g}\)). 1 mark: Method to find mass of copper or zinc in Alloy B (e.g., \(420\text{ g}\) or \(180\text{ g}\)). 1.6 marks: Correctly calculating total masses of \(640\text{ g}\) and \(360\text{ g}\) and simplifying to \(16 : 9\).

Let the original price of the coat be \(P\). The first reduction of \(20\%\) means the price is multiplied by \(0.80\). The second reduction of \(15\%\) means this new price is multiplied by \(0.85\). Therefore, \(P \times 0.80 \times 0.85 = 54.40\). Since \(0.80 \times 0.85 = 0.68\), we have \(0.68P = 54.40\). Solving for \(P\): \(P = \frac{54.40}{0.68} = 80\). The original price was £\(80\).

1.5 marks: Setting up an equation or finding the combined multiplier, e.g., \(0.80 \times 0.85 = 0.68\) (or equivalent percentage representation of \(68\%\)). 1 mark: Method to divide the final clearance price by the combined multiplier, e.g., \(54.40 \div 0.68\). 1.1 marks: Correct final answer of \(80\).

The area of a rectangle is length multiplied by width, so: \((x+5)(2x-3) = 75\). Expand the brackets: \(2x^2 - 3x + 10x - 15 = 75\), which simplifies to \(2x^2 + 7x - 15 = 75\). Subtract 75 from both sides to form a quadratic equation equal to zero: \(2x^2 + 7x - 90 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 2\), \(b = 7\), and \(c = -90\): \(x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-90)}}{2(2)}\), which simplifies to \(x = \frac{-7 \pm \sqrt{49 + 720}}{4} = \frac{-7 \pm \sqrt{769}}{4}\). Calculating this gives \(x \approx \frac{-7 + 27.7308}{4} \approx 5.1827\) or a negative value which is rejected since width and length must be positive. Therefore, to 3 significant figures, \(x = 5.18\).

1 mark: Setting up the initial equation and correctly expanding/rearranging to the quadratic form \(2x^2 + 7x - 90 = 0\). 1.5 marks: Correctly substituting values into the quadratic formula. 1.1 marks: Finding the correct positive solution rounded to 3 significant figures (\(5.18\)).

The total volume is the sum of the volume of the hemisphere and the volume of the cone. Volume of hemisphere = \(\frac{2}{3}\pi r^3\). Volume of cone = \(\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3\). Total Volume = \(\frac{2}{3}\pi r^3 + \frac{2}{3}\pi r^3 = \frac{4}{3}\pi r^3\). Given the total volume is \(288\pi\text{ cm}^3\): \(\frac{4}{3}\pi r^3 = 288\pi\). Dividing both sides by \(\pi\) gives \(\frac{4}{3}r^3 = 288\). Multiplying both sides by \(\frac{3}{4}\) gives \(r^3 = 216\). Taking the cube root of both sides gives \(r = \sqrt[3]{216} = 6\text{ cm}\).

1 mark: Correct algebraic expressions for both volumes in terms of \(r\), e.g., \(\frac{2}{3}\pi r^3\) for the hemisphere and \(\frac{2}{3}\pi r^3\) for the cone. 1 mark: Equating the total combined volume \(\frac{4}{3}\pi r^3\) to \(288\pi\). 1.6 marks: Correctly solving for \(r = 6\).

Let the number of red counters be \(3x\) and the number of blue counters be \(2x\). The total number of counters is \(5x\). The probability of selecting a red counter first is \(\frac{3x}{5x} = \frac{3}{5}\). Since the choice is without replacement, the probability of selecting a red counter second is \(\frac{3x - 1}{5x - 1}\). The probability of both being red is: \(\frac{3}{5} \times \frac{3x - 1}{5x - 1} = \frac{1}{3}\). This simplifies to \(\frac{9x - 3}{25x - 5} = \frac{1}{3}\). Cross-multiplying gives \(3(9x - 3) = 25x - 5\), which is \(27x - 9 = 25x - 5\). Rearranging gives \(2x = 4\), so \(x = 2\). The total number of counters is \(5x = 5(2) = 10\).

1.5 marks: Formulating the probability equation, e.g., \(\frac{3}{5} \times \frac{3x - 1}{5x - 1} =
\frac{1}{3}\). 1 mark: Correctly expanding and simplifying to a linear equation in \(x\), e.g., \(27x - 9 = 25x - 5\). 1.1 marks: Correctly solving for \(x = 2\) and stating the total number of counters is \(10\).

Let the interior angle be \(I\) and the exterior angle be \(E\). We know that the sum of the interior and exterior angles is always \(180^\circ\), so \(I + E = 180\). We are also given that \(I = E + 140\). Substituting the second equation into the first gives: \((E + 140) + E = 180\), which simplifies to \(2E + 140 = 180\). Subtracting 140 from both sides gives \(2E = 40\), so \(E = 20^\circ\). The number of sides \(n\) of a regular polygon is given by \(n = \frac{360^\circ}{\text{exterior angle}}\). Thus, \(n = \frac{360}{20} = 18\).

1 mark: Setting up simultaneous equations or algebraic expression relating interior and exterior angles (e.g., \(I + E = 180\) and \(I - E = 140\)). 1 mark: Finding the correct exterior angle of \(20^\circ\). 1.6 marks: Dividing \(360^\circ\) by \(20^\circ\) to obtain the correct final answer of \(18\).

First, find the lower and upper bounds for distance and time. Distance \(d\) is 120 correct to the nearest 10, so the upper bound is \(d_{\text{max}} = 125\text{ km}\). Time \(t\) is 2.4 correct to 1 decimal place (nearest 0.1), so the lower bound is \(t_{\text{min}} = 2.35\text{ hours}\). To find the upper bound for average speed, use the formula: \(\text{Speed}_{\text{max}} = \frac{d_{\text{max}}}{t_{\text{min}}} = \frac{125}{2.35} \approx 53.191489...\text{ km/h}\). Rounding this to 1 decimal place gives \(53.2\text{ km/h}\).

1.5 marks: Stating the correct bounds for distance (\(d_{\text{max}} = 125\)) and time (\(t_{\text{min}} = 2.35\)). 1 mark: Correct calculation of speed upper bound, i.e., \(\frac{125}{2.35}\). 1.1 marks: Providing the final answer correct to 1 decimal place (\(53.2\)).

Find the differences between consecutive terms: 1st differences: \(11\), \(17\), \(23\). 2nd differences: \(6\), \(6\). Because the second difference is constant and equal to 6, the coefficient of the \(n^2\) term is \(a = \frac{6}{2} = 3\). Subtract \(3n^2\) from each term of the original sequence to find the linear part: For \(n=1\): \(4 - 3(1)^2 = 1\). For \(n=2\): \(15 - 3(2)^2 = 3\). For \(n=3\): \(32 - 3(3)^2 = 5\). For \(n=4\): \(55 - 3(4)^2 = 7\). The sequence of differences \(1, 3, 5, 7, \dots\) has the formula \(2n - 1\). Thus, the \(n\)-th term of the quadratic sequence is \(3n^2 + 2n - 1\). Substituting \(n = 20\): \(3(20)^2 + 2(20) - 1 = 3(400) + 40 - 1 = 1200 + 40 - 1 = 1239\).

1 mark: Finding the constant second difference of 6 and identifying that the term in \(n^2\) is \(3n^2\). 1.5 marks: Correctly finding the complete \(n\)-th term formula: \(3n^2 + 2n - 1\). 1.1 marks: Substituting \(n = 20\) to find the final answer \(1239\).

First, find the volume of copper and zinc in the alloy.

Total parts in the ratio = \(3 + 2 = 5\)

Volume of copper = \(\frac{3}{5} \times 350 = 210\text{ cm}^3\)
Volume of zinc = \(\frac{2}{5} \times 350 = 140\text{ cm}^3\)

Next, calculate the mass of each metal using the formula \(\text{Mass} = \text{Density} \times \text{Volume}\):

Mass of copper = \(8.9 \times 210 = 1869\text{ g}\)
Mass of zinc = \(7.1 \times 140 = 994\text{ g}\)

Finally, add the masses together to find the total mass:

Total mass = \(1869 + 994 = 2863\text{ g}\)

M1: For finding the volume of copper or zinc (\(210\) or \(140\)) or showing a correct method to split the volume, e.g., \(\frac{3}{5} \times 350\) or \(\frac{2}{5} \times 350\)
M1: For a correct method to find the mass of either the copper or the zinc, e.g., \(\text{their } 210 \times 8.9\) or \(\text{their } 140 \times 7.1\)
M1: For a complete method to calculate both masses and add them together
A1: \(2863\)

Let the original value of the company be \(V\).

We can express the changes over the three years using multipliers:
- Year 1 increase of \(12\%\) corresponds to a multiplier of \(1.12\)
- Year 2 decrease of \(5\%\) corresponds to a multiplier of \(0.95\)
- Year 3 increase of \(8\%\) corresponds to a multiplier of \(1.08\)

Set up the equation for the final value:
\(V \times 1.12 \times 0.95 \times 1.08 = 287280\)

Calculate the combined multiplier:
\(1.12 \times 0.95 \times 1.08 = 1.14912\)

Solve for \(V\):
\(V = \frac{287280}{1.14912} = 250000\)

The original value of the company was \(\text{£}250,000\).

M1: For identifying any correct single multiplier (e.g., \(1.12\), \(0.95\), or \(1.08\)) or attempting to work backward by one year (e.g., \(\frac{287280}{1.08} = 266000\))
M1: For a correct method to find the combined multiplier, e.g., \(1.12 \times 0.95 \times 1.08\) (\(= 1.14912\)), or working backward by two years, e.g., \(\frac{266000}{0.95} = 280000\)
M1: For a complete division calculation to find the original value, e.g., \(\frac{287280}{1.14912}\) or \(\frac{280000}{1.12}\)
A1: \(250000\) (accept \(\text{£}250,000\))

Write down the formula for the volume of a cylinder:
\(V_{\text{cylinder}} = \pi r^2 h\)

Substitute \(h = 3r\):
\(V_{\text{cylinder}} = \pi \times r^2 \times 3r = 3\pi r^3\)

Write down the formula for the volume of a sphere:
\(V_{\text{sphere}} = \frac{4}{3}\pi r^3\)

Add the two volumes together to get the total combined volume:
\(\text{Total Volume} = 3\pi r^3 + \frac{4}{3}\pi r^3 = \frac{13}{3}\pi r^3\)

Set this equal to the given total volume:
\(\frac{13}{3}\pi r^3 = 936\pi\)

Divide both sides by \(\pi\):
\(\frac{13}{3}r^3 = 936\)

Multiply both sides by \(3\) and divide by \(13\):
\(r^3 = 936 \times \frac{3}{13}\)
\(r^3 = 72 \times 3\)
\(r^3 = 216\)

Take the cube root of both sides:
\(r = \sqrt[3]{216} = 6\)

M1: For expressing the volume of the cylinder as \(3\pi r^3\) or the volume of the sphere as \(\frac{4}{3}\pi r^3\)
M1: For setting up the equation \(3\pi r^3 + \frac{4}{3}\pi r^3 = 936\pi\) (or equivalent without \(\pi\))
M1: For simplifying the equation to find \(r^3 = 216\)
A1: \(6\)