2024 AQA GCSE Mathematics 8300 Practice Paper with Answers

First, work out the calculation inside the brackets:
\( 4 - 7 = -3 \)

Next, square this result:
\( (-3)^2 = 9 \)

Then, multiply by 3:
\( 3 \times 9 = 27 \)

Finally, subtract this from 15:
\( 15 - 27 = -12 \)

M1 for correctly calculating \( (4 - 7)^2 = 9 \) or writing \( 15 - 3 \times 9 \)
A0.5 for \( -12 \)

First, expand the brackets:
\( 4(2x - 3) = 8x - 12 \)
\( -3(x - 5) = -3x + 15 \)

Next, collect like terms:
\( 8x - 3x - 12 + 15 = 5x + 3 \)

M1 for expanding at least one bracket correctly to get \( 8x - 12 \) or \( -3x + 15 \)
A0.5 for \( 5x + 3 \)

First, find the actual distance in centimetres by multiplying by the scale factor:
\( 6\text{ cm} \times 25\ 000 = 150\ 000\text{ cm} \)

Convert centimetres to metres by dividing by 100:
\( 150\ 000 \div 100 = 1500\text{ m} \)

Convert metres to kilometres by dividing by 1000:
\( 1500 \div 1000 = 1.5\text{ km} \)

M1 for a correct conversion method, such as showing \( 6 \times 25\ 000 = 150\ 000 \) or converting the scale to \( 1\text{ cm} = 0.25\text{ km} \)
A0.5 for \( 1.5 \) (accept \( 1.5\text{ km} \))

First, convert the mixed number to an improper fraction:
\( 1\frac{3}{5} = \frac{8}{5} \)

Next, multiply by the reciprocal of the divisor:
\( \frac{8}{5} \times \frac{3}{2} = \frac{24}{10} \)

Simplify the fraction:
\( \frac{24}{10} = \frac{12}{5} \)

Convert the improper fraction to a mixed number:
\( \frac{12}{5} = 2\frac{2}{5} \)

M1 for converting to \( \frac{8}{5} \) and multiplying by \( \frac{3}{2} \) to obtain \( \frac{24}{10} \) or \( \frac{12}{5} \)
A0.5 for \( 2\frac{2}{5} \)

Subtract 5 from both sides of the inequality:
\( -2x < 6 \)

Divide both sides by -2 and reverse the inequality sign because you are dividing by a negative number:
\( x > -3 \)

M1 for subtracting 5 from both sides to get \( -2x < 6 \) (or rearranging to \( 2x > -6 \))
A0.5 for \( x > -3 \)

The sum of all probabilities must equal 1.

\( \text{Probability of yellow} = 1 - (0.35 + 0.4) \)
\( = 1 - 0.75 \)
\( = 0.25 \)

M1 for adding the given probabilities to get \( 0.75 \) or writing \( 1 - 0.35 - 0.4 \)
A0.5 for \( 0.25 \) (or equivalent fraction / percentage)

The measurement is correct to 1 decimal place, which is to the nearest \( 0.1\text{ cm} \).
Half of this degree of accuracy is:
\( 0.1 \div 2 = 0.05\text{ cm} \)

To find the upper bound, add \( 0.05\text{ cm} \) to the measured value:
\( 8.4 + 0.05 = 8.45\text{ cm} \)

M1 for identifying \( 0.05 \) as the boundary increment, or stating the lower bound as \( 8.35 \)
A0.5 for \( 8.45 \) (accept \( 8.45\text{ cm} \))

Find the difference between consecutive terms:
\( 11 - 7 = 4 \)
\( 15 - 11 = 4 \)

Since the common difference is 4, the formula contains \( 4n \).

Compare \( 4n \) with the sequence:
When \( n = 1 \), \( 4(1) = 4 \), but we need 7 (add 3).
When \( n = 2 \), \( 4(2) = 8 \), but we need 11 (add 3).

Therefore, the \( n \)-th term is \( 4n + 3 \).

M1 for identifying a common difference of 4 or writing \( 4n \)
A0.5 for \( 4n + 3 \)

Let \(x = 0.1\dot{7} = 0.1777...\)
Then \(10x = 1.777...\)
And \(100x = 17.777...\)
Subtracting the two equations:
\(100x - 10x = 17.777... - 1.777...\)
\(90x = 16\)
\(x = \frac{16}{90} = \frac{8}{45}\).

M1: For setting up two correct equations that would eliminate the recurring decimal part when subtracted, such as \(10x = 1.77...\) and \(100x = 17.77...\) (or equivalent).
A1 (0.5 marks): For \(\frac{8}{45}\) (must be in simplest form).

Multiply both sides by 3:
\(2x + 7 = 3(x - 1)\)
\(2x + 7 = 3x - 3\)
Subtract \(2x\) from both sides:
\(7 = x - 3\)
Add 3 to both sides:
\(x = 10\).

M1: For removing the fraction by multiplying both sides by 3, resulting in \(2x + 7 = 3(x - 1)\) or \(2x + 7 = 3x - 3\)
A1 (0.5 marks): For \(10\) (or \(x = 10\)).

Since \(y\) is inversely proportional to the square of \(x\), we can write:
\(y = \frac{k}{x^2}\)
Substitute \(x = 3\) and \(y = 4\) to find \(k\):
\(4 = \frac{k}{3^2}\)
\(4 = \frac{k}{9} \implies k = 36\)
So, the formula is:
\(y = \frac{36}{x^2}\)
When \(x = 6\):
\(y = \frac{36}{6^2} = \frac{36}{36} = 1\).

M1: For establishing the equation \(y = \frac{k}{x^2}\) and substituting values to find the constant of proportionality \(k = 36\) (or showing that \(y_1 \cdot x_1^2 = y_2 \cdot x_2^2\)).
A1 (0.5 marks): For \(1\).

We look for two numbers that multiply to \(2 \times (-3) = -6\) and add up to \(5\). These numbers are \(6\) and \(-1\).
Rewrite the middle term:
\(2x^2 + 6x - x - 3\)
Factorise by grouping:
\(2x(x + 3) - 1(x + 3)\)
This gives:
\((2x - 1)(x + 3)\).

M1: For any factorisation that expands to give at least two of the three terms correct (e.g., \((2x - 3)(x + 1)\) or showing the split of the middle term as \(6x - x\)).
A1 (0.5 marks): For \((2x - 1)(x + 3)\) or \((x + 3)(2x - 1)\).

Let \(H\) be History and \(G\) be Geography.
The number of students studying only History is \(22 - 8 = 14\).
The number of students studying only Geography is \(15 - 8 = 7\).
The number of students studying at least one of these subjects is:
\(14 + 7 + 8 = 29\)
Therefore, the number of students who study neither subject is:
\(40 - 29 = 11\)
The probability of choosing a student who studies neither is \(\frac{11}{40}\).

M1: For calculating the number of students who study neither subject as \(11\) (e.g., showing \(40 - (22 + 15 - 8)\)).
A1 (0.5 marks): For \(\frac{11}{40}\) (or equivalent fraction, decimal \(0.275\), or percentage \(27.5\%\)).

First, find the area of the triangular cross-section:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 30\text{ cm}^2\)
Then, find the volume of the prism by multiplying the cross-sectional area by its length:
\(\text{Volume} = \text{Area} \times \text{length} = 30 \times 15 = 450\text{ cm}^3\).

M1: For calculating the cross-sectional area of the triangle as \(30\text{ cm}^2\) (e.g., \(\frac{1}{2} \times 5 \times 12\)).
A1 (0.5 marks): For \(450\).

We can rewrite \(27^{-\frac{2}{3}}\) as:
\(\frac{1}{27^{\frac{2}{3}}}\)
First, find the cube root of 27:
\(27^{\frac{1}{3}} = \sqrt[3]{27} = 3\)
Next, square this result:
\(3^2 = 9\)
Therefore, we have:
\(\frac{1}{9}\).

M1: For applying either the negative index or fractional index correctly (e.g., showing \(3^{-2}\) or \(\frac{1}{27^{2/3}}\) or \(\frac{1}{9}\) without full working).
A1 (0.5 marks): For \(\frac{1}{9}\).

Find the first differences between consecutive terms:
5, 7, 9, 11
Find the second differences:
2, 2, 2
Since the second difference is constant at 2, the coefficient of the \(n^2\) term is \(\frac{2}{2} = 1\). So the sequence contains \(n^2\).
Subtract \(n^2\) from each term of the original sequence:
Position \(n\): 1, 2, 3, 4, 5
Sequence: 3, 8, 15, 24, 35
\(n^2\): 1, 4, 9, 16, 25
Difference: 2, 4, 6, 8, 10
The remaining sequence is \(2, 4, 6, 8, 10\), which has an \(n\)-th term of \(2n\).
Combining these gives the final \(n\)-th term:
\(n^2 + 2n\).

M1: For finding the second difference of 2 and writing down \(n^2\) as part of the formula, or showing the linear sequence \(2, 4, 6, 8, 10\) after subtracting \(n^2\).
A1 (0.5 marks): For \(n^2 + 2n\) (or equivalent expression).

1. Find the number of green marbles: Since the ratio of blue to green is \(2 : 1\) and there are 24 blue marbles, there are \(24 \div 2 = 12\) green marbles. 2. Find the total number of blue and green marbles: \(24 + 12 = 36\) marbles. 3. Since \(\frac{2}{5}\) of the marbles are red, the remaining \(\frac{3}{5}\) must be blue and green. 4. Let \(T\) be the total number of marbles: \(\frac{3}{5}T = 36 \implies T = 36 \times \frac{5}{3} = 12 \times 5 = 60\).

M1 for finding the number of green marbles as 12 (or total blue and green as 36). A0.5 for the final correct answer 60.

1. Subtract 1 from all parts of the inequality: \(-3 - 1 < 2x \le 7 - 1\) which simplifies to \(-4 < 2x \le 6\). 2. Divide all parts by 2: \(-2 < x \le 3\). 3. Identify the integers in this range: since \(x\) is strictly greater than \(-2\) and less than or equal to \(3\), the integers are \(-1, 0, 1, 2, 3\).

M1 for isolating \(x\) to find the interval \(-2 < x \le 3\) (or list showing at least 3 correct integers with no more than one incorrect). A0.5 for the complete correct list of integers: -1, 0, 1, 2, 3.

1. Find the exterior angle of the regular polygon: Since the interior and exterior angles on a straight line add up to \(180^\circ\), the exterior angle is \(180^\circ - 162^\circ = 18^\circ\). 2. Use the sum of exterior angles (\(360^\circ\)) to find the number of sides \(n\): \(n = 360^\circ \div 18^\circ = 20\).

M1 for finding the exterior angle as \(18^\circ\) or writing a correct equation such as \(\frac{(n-2)\times 180}{n} = 162\). A0.5 for 20.

1. Factorise the numerator \(6x^2 - 13x - 5\): We look for two numbers that multiply to \(6 \times -5 = -30\) and add to \(-13\). These are \(-15\) and \(2\). Grouping gives: \(6x^2 - 15x + 2x - 5 = 3x(2x - 5) + 1(2x - 5) = (3x + 1)(2x - 5)\). 2. Factorise the denominator \(4x^2 - 25\): This is a difference of two squares: \((2x - 5)(2x + 5)\). 3. Simplify by cancelling the common factor \((2x - 5)\): \(\frac{(3x + 1)(2x - 5)}{(2x - 5)(2x + 5)} = \frac{3x + 1}{2x + 5}\).

M1 for factorising either the numerator to \((3x + 1)(2x - 5)\) or the denominator to \((2x - 5)(2x + 5)\). A0.5 for the fully simplified expression \(\frac{3x + 1}{2x + 5}\).

1. Calculate the probability of picking two red counters: \(P(\text{Red, Red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}\). 2. Calculate the probability of picking two blue counters: \(P(\text{Blue, Blue}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}\). 3. Add these probabilities together to find the probability of same colour: \(P(\text{same colour}) = \frac{20}{56} + \frac{6}{56} = \frac{26}{56}\). 4. Simplify the fraction: \(\frac{26}{56} = \frac{13}{28}\).

M1 for finding either \(P(\text{Red, Red}) = \frac{20}{56}\) or \(P(\text{Blue, Blue}) = \frac{6}{56}\) (or equivalents). A0.5 for \(\frac{13}{28}\) (or any equivalent fraction/decimal).

Convert both numbers to ordinary numbers or to the same power of 10. \(4 \times 10^5 = 400,000\) and \(6 \times 10^4 = 60,000\). Subtract the two values: \(400,000 - 60,000 = 340,000\). Convert the result back to standard form: \(340,000 = 3.4 \times 10^5\).

M1 for converting to ordinary numbers or to a common power of 10, e.g., \(400,000\) and \(60,000\) or \(40 \times 10^4 - 6 \times 10^4\). M1 for subtraction leading to \(340,000\) or \(34 \times 10^4\). A1 for the correct standard form answer \(3.4 \times 10^5\).

Let \(x = 0.2454545...\). Multiply by 10 to get \(10x = 2.454545...\). Multiply by 1000 to get \(1000x = 245.454545...\). Subtract the two equations: \(1000x - 10x = 245.454545... - 2.454545...\) which gives \(990x = 243\). Solve for \(x\): \(x = \frac{243}{990}\). Simplify the fraction by dividing both numerator and denominator by 9: \(243 \div 9 = 27\) and \(990 \div 9 = 110\). Thus, \(x = \frac{27}{110}\).

M1 for establishing two equations that can be subtracted to eliminate the recurring part, e.g., \(10x = 2.4545...\) and \(1000x = 245.4545...\). M1 for subtracting to get a correct equation such as \(990x = 243\). A1 for simplifying \(\frac{243}{990}\) to \(\frac{27}{110}\) with all steps shown clearly.

Find 10% of £12 000: \(0.10 \times 12000 = 1200\). Calculate the value of the car after the first year: \(12000 - 1200 = 10800\). Find 20% of £10 800: \(0.20 \times 10800 = 2160\). Calculate the value of the car after the second year: \(10800 - 2160 = 8640\). The final value of the car is £8640.

M1 for finding the value of the car after the 10% depreciation, i.e., \(12000 - 1200 = 10800\). M1 for finding 20% of their £10 800 (\(2160\)) and subtracting it from their £10 800. A1 for £8640.

Substitute the first equation into the second: \(x^2 + (2x + 1)^2 = 10\). Expand the brackets: \(x^2 + 4x^2 + 4x + 1 = 10\). Simplify and set to zero: \(5x^2 + 4x - 9 = 0\). Factorise the quadratic equation: \((5x + 9)(x - 1) = 0\). Solve for \(x\): \(5x + 9 = 0 \implies x = -1.8\) or \(x - 1 = 0 \implies x = 1\).

M1 for substituting \(y = 2x + 1\) into the second equation and expanding, getting \(x^2 + 4x^2 + 4x + 1 = 10\). M1 for setting to quadratic form \(5x^2 + 4x - 9 = 0\) and attempting to factorise or use formula. A1 for finding both values: \(x = 1\) and \(x = -1.8\) (or \(-\frac{9}{5}\)).

Find the first differences: \(10 - 3 = 7\), \(21 - 10 = 11\), and \(36 - 21 = 15\). Find the second differences: \(11 - 7 = 4\) and \(15 - 11 = 4\). Since the second difference is constant at 4, the coefficient of the \(n^2\) term is \(4 \div 2 = 2\). Subtract \(2n^2\) from each term in the sequence: for \(n=1\), \(3 - 2 = 1\); for \(n=2\), \(10 - 8 = 2\); for \(n=3\), \(21 - 18 = 3\); for \(n=4\), \(36 - 32 = 4\). The remaining sequence is 1, 2, 3, 4, which has the general formula \(n\). Combining these gives the \(n\)-th term: \(2n^2 + n\).

M1 for finding the second differences (4) and identifying that the \(n^2\) term has a coefficient of 2 (i.e., \(2n^2\)). M1 for subtracting \(2n^2\) from the terms to get the linear remainder sequence 1, 2, 3, 4. A1 for the complete correct expression \(2n^2 + n\).

There are two ways to get different colours: Red then Blue (RB) or Blue then Red (BR). Calculate \(P(\text{RB})\): \(P(\text{Red first}) = \frac{5}{8}\), \(P(\text{Blue second}) = \frac{3}{7}\), so \(P(\text{RB}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\). Calculate \(P(\text{BR})\): \(P(\text{Blue first}) = \frac{3}{8}\), \(P(\text{Red second}) = \frac{5}{7}\), so \(P(\text{BR}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\). Add the two probabilities: \(P(\text{different colours}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\).

M1 for calculating the probability of one valid combination, e.g., \(\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\). M1 for summing the two correct probabilities \(\frac{15}{56} + \frac{15}{56}\). A1 for \(\frac{15}{28}\) or equivalent simplified fraction.

Calculate the volume of the cylinder: \(V = \pi \times r^2 \times h = \pi \times 2^2 \times 9 = 36\pi\). Set the volume of the sphere equal to the volume of the cylinder: \(\frac{4}{3}\pi r^3 = 36\pi\). Divide both sides by \(\pi\): \(\frac{4}{3} r^3 = 36\). Multiply by \(\frac{3}{4}\): \(r^3 = 27\). Take the cube root: \(r = 3\text{ cm}\).

M1 for finding the volume of the cylinder as \(36\pi\). M1 for setting up the equation \(\frac{4}{3}\pi r^3 = 36\pi\) and rearranging to find \(r^3 = 27\). A1 for \(3\text{ cm}\) (units not strictly required for accuracy mark but must show 3).

Find the midpoint of \(AB\): \(x = \frac{2 + 6}{2} = 4\) and \(y = \frac{5 + 13}{2} = 9\), giving midpoint \((4, 9)\). Find the gradient of \(L_1\): \(m_1 = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2\). Find the gradient of the perpendicular line \(L_2\): \(m_2 = -\frac{1}{2}\). Find the equation of \(L_2\) using \(y - y_1 = m(x - x_1)\): \(y - 9 = -\frac{1}{2}(x - 4)\) which simplifies to \(y = -\frac{1}{2}x + 11\).

M1 for finding the midpoint of \(AB\) as \((4, 9)\). M1 for finding the gradient of \(L_1\) as 2 and determining the perpendicular gradient is \(-\frac{1}{2}\). A1 for the correct equation in the requested form, e.g., \(y = -\frac{1}{2}x + 11\).

To find the value of \(w\), express each base as a power of 3: \(27 = 3^3\), so \(27^4 = (3^3)^4 = 3^{12}\) and \(9 = 3^2\), so \(9^3 = (3^2)^3 = 3^6\). Now substitute these into the original equation: \(\frac{3^{12} \times 3^5}{3^6} = 3^w\). Simplify the numerator using the rule \(a^m \times a^n = a^{m+n}\): \(3^{12} \times 3^5 = 3^{12+5} = 3^{17}\). Now simplify the fraction using the rule \(\frac{a^m}{a^n} = a^{m-n}\): \(\frac{3^{17}}{3^6} = 3^{17-6} = 3^{11}\). Therefore, \(3^{11} = 3^w\), which gives \(w = 11\).

M1: Expressing 27 as \(3^3\) or 9 as \(3^2\) or writing \(27^4\) as \(3^{12}\) or \(9^3\) as \(3^6\).
M1: Applying laws of indices to simplify the expression to a single power of 3 (e.g., finding \(3^{17}\) in the numerator or carrying out subtraction of powers to get \(3^{11}\)).
A1: Correct value of \(w = 11\).

The ratio of their shares is: Alice: 3 parts, Bob: 5 parts, Charlie: 8 parts. The difference in parts between Charlie and Alice is \(8 - 3 = 5\text{ parts}\). We are told that Charlie receives 120 more than Alice, so 5 parts correspond to 120: \(5\text{ parts} = 120\). Therefore, \(1\text{ part} = \frac{120}{5} = 24\). The total number of parts shared is \(3 + 5 + 8 = 16\text{ parts}\). The total amount of money is \(16 \times 24 = 384\).

M1: Finds the difference in ratio parts between Charlie and Alice: \(8 - 3 = 5\) parts.
M1: Calculates the value of one part by doing \(\frac{120}{5} = 24\) or sets up a correct equation like \(\frac{3}{16}x + 120 = \frac{8}{16}x\).
A1: Correct total amount of 384.

Let \(s\) be the price of a sourdough loaf in pounds, and \(b\) be the price of a baguette in pounds. We can set up two simultaneous equations: (1) \(2s + 3b = 11.50\) and (2) \(3s + 2b = 13.50\). Multiply equation (1) by 2 to get \(4s + 6b = 23.00\) (3). Multiply equation (2) by 3 to get \(9s + 6b = 40.50\) (4). Subtract equation (3) from equation (4): \(5s = 17.50\), which gives \(s = 3.50\). Substitute \(s = 3.50\) back into equation (1): \(2(3.50) + 3b = 11.50\), which simplifies to \(7.00 + 3b = 11.50\), so \(3b = 4.50\) and \(b = 1.50\). We want to find the total cost of 1 sourdough loaf and 1 baguette: \(s + b = 3.50 + 1.50 = 5.00\).

M1: Formulates two correct simultaneous equations, e.g., \(2s + 3b = 11.5\) and \(3s + 2b = 13.5\), and attempts to eliminate one variable by multiplying the equations.
M1: Successfully solves for one variable, finding \(s = 3.5\) or \(b = 1.5\).
A1: Finds the correct total cost of 5.00 (accept 5 or 5.00).

To find the volume of a prism, we use the formula: \(\text{Volume} = \text{Area of cross-section} \times \text{length}\). First, we find the height, \(h\), of the right-angled triangular cross-section using Pythagoras' theorem: \(h^2 + 5^2 = 13^2\), so \(h^2 + 25 = 169\), which gives \(h^2 = 144\) and \(h = 12\text{ cm}\). Now, calculate the area of the triangular cross-section: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 30\text{ cm}^2\). Finally, calculate the volume of the prism: \(\text{Volume} = 30\text{ cm}^2 \times 12\text{ cm} = 360\text{ cm}^3\).

M1: Uses Pythagoras' theorem to find the missing side of the triangle: \(\sqrt{13^2 - 5^2} = 12\).
M1: Calculates the cross-sectional area of the triangle: \(\frac{1}{2} \times 5 \times 12 = 30\).
A1: Correct volume of 360.

Divide the coefficients and subtract the powers of 10: \( 2.4 \div 3 = 0.8 \) and \( 10^5 \div 10^{-2} = 10^{5 - (-2)} = 10^7 \). This gives \( 0.8 \times 10^7 \). In standard form, this is written as \( 8 \times 10^6 \).

M1 for finding either the coefficient \( 0.8 \) or the power \( 10^7 \). A1 for the correct standard form \( 8 \times 10^6 \).

Let \( x = 0.1777... \). Then \( 10x = 1.777... \) and \( 100x = 17.777... \). Subtracting the first equation from the second gives \( 90x = 16 \). Solving for \( x \) gives \( x = \frac{16}{90} \), which simplifies to \( \frac{8}{45} \).

M1 for attempting to subtract two equations to eliminate the recurring part, e.g. \( 100x - 10x = 17.77... - 1.77... \) to obtain \( 90x = 16 \). A1 for simplifying \( \frac{16}{90} \) to \( \frac{8}{45} \).

The highest common factor of the numerical coefficients \( 12 \) 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 \( y \). Thus, the common factor to divide out is \( 6xy \). Dividing both terms by \( 6xy \) leaves \( 2x - 3y \). The fully factorised expression is \( 6xy(2x - 3y) \).

M1 for a partially factorised expression, such as extracting a common factor of at least \( 6x \), \( 6y \), or \( xy \). A1 for the fully correct factorised form \( 6xy(2x - 3y) \).

Since the probabilities of all outcomes must add up to 1, we have \( P(\text{red}) + P(\text{blue}) + P(\text{green}) = 1 \). Substituting the given values: \( 0.4 + 2P(\text{green}) + P(\text{green}) = 1 \). This simplifies to \( 0.4 + 3P(\text{green}) = 1 \), which means \( 3P(\text{green}) = 0.6 \). Dividing by 3 gives \( P(\text{green}) = 0.2 \).

M1 for setting up the equation \( 3x = 1 - 0.4 \) or equivalent. A1 for the correct probability \( 0.2 \).

The formula for the volume of a cylinder is \( V = \pi r^2 h \). Substituting the radius \( r = 5 \) and height \( h = 8 \) gives \( V = \pi \times 5^2 \times 8 = \pi \times 25 \times 8 = 200\pi\text{ cm}^3 \).

M1 for substituting \( r = 5 \) and \( h = 8 \) into the cylinder volume formula. A1 for obtaining \( 200\pi \).

Let the volume of the standard size bottle be \(x\) ml.

Since 450 ml is 25% more than the standard size, we can write the equation:
\(1.25x = 450\)

To find \(x\), divide both sides by 1.25:
\(x = \frac{450}{1.25} = 360\) ml.

M1 for \(450 \div 1.25\) or for setting up the equation \(1.25x = 450\) (or equivalent).
A0.5 for 360.

First, simplify the numerator:
\(4.8 \times 10^5 + 3.2 \times 10^4 = 480,000 + 32,000 = 512,000\)
In standard form, the numerator is \(5.12 \times 10^5\).

Now, divide by the denominator:
\(\frac{5.12 \times 10^5}{1.6 \times 10^{-2}} = \left(\frac{5.12}{1.6}\right) \times 10^{5 - (-2)}\)
\(= 3.2 \times 10^7\)

M1 for finding the numerator as \(512,000\) or \(5.12 \times 10^5\), or for showing division of the coefficients.
A0.5 for \(3.2 \times 10^7\).

First, convert the mass from kilograms to grams:
\(1.2\text{ kg} = 1.2 \times 1000 = 1200\text{ g}\)

Next, use the formula for density:
\(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\)
\(\text{Density} = \frac{1200}{150} = 8\text{ g/cm}^3\)

M1 for converting mass to \(1200\text{ g}\) or using the density formula correctly with units.
A0.5 for 8.

The formula for the volume of a cylinder is:
\(V = \pi r^2 h\)

Substitute the given values:
\(V = \pi \times 4^2 \times 10\)
\(V = \pi \times 16 \times 10\)
\(V = 160\pi \approx 502.65482\dots\text{ cm}^3\)

Rounding to 1 decimal place gives \(502.7\text{ cm}^3\).

M1 for \(\pi \times 4^2 \times 10\) or \(160\pi\).
A0.5 for 502.7.

First, expand the bracket on the left-hand side:
\(8x - 12 = 3x + 13\)

Subtract \(3x\) from both sides:
\(5x - 12 = 13\)

Add 12 to both sides:
\(5x = 25\)

Divide by 5:
\(x = 5\)

M1 for expanding the bracket correctly to \(8x - 12\) or for isolating the \(x\) terms on one side of the equation.
A0.5 for 5.

First, find the probability that the coin lands on Tails:
\(P(\text{Tails}) = 1 - 0.65 = 0.35\)

Now, calculate the expected number of times it lands on Tails over 200 spins:
\(\text{Expected value} = 200 \times 0.35 = 70\)

M1 for calculating \(1 - 0.65\) or for calculating the expected number of Heads as \(200 \times 0.65 = 130\).
A0.5 for 70.

Since the mean of 5 numbers is 12, their sum is:
\(5 \times 12 = 60\)

The sum of the four known numbers is:
\(8 + 15 + 11 + 14 = 48\)

The fifth number is the difference:
\(60 - 48 = 12\)

M1 for \(5 \times 12\) or \(8 + 15 + 11 + 14\).
A0.5 for 12.

First, expand both sets of brackets:
\(3(2x + 5y) = 6x + 15y\)

\(-2(x - 4y) = -2x + 8y\)

Now combine the expanded terms:
\(6x + 15y - 2x + 8y\)

Collect like terms:
\(6x - 2x = 4x\)

\(15y + 8y = 23y\)

So the simplified expression is \(4x + 23y\).

M1 for correct expansion of at least one bracket (e.g. \(6x + 15y\) or \(-2x + 8y\)).
A0.5 for \(4x + 23y\).

The sale price represents \(100\% - 15\% = 85\%\) of the original price. To find the original price, we calculate: \(442 \div 0.85 = 520\). Therefore, the original price is \(£520\).

M1 for \(442 \div 0.85\) or equivalent percentage method. A0.5 for \(520\) (accept £520).

Since there are 42 blue counters which correspond to 7 parts of the ratio, each part is equal to \(42 \div 7 = 6\) counters. The number of red counters corresponds to 3 parts: \(3 \times 6 = 18\).

M1 for \(42 \div 7 \times 3\) or equivalent complete method. A0.5 for 18.

First, add 3 to both sides of the inequality to isolate the term with \(x\): \(5x > 25\). Next, divide both sides by 5: \(x > 5\).

M1 for rearranging to get \(5x > 25\) or showing addition of 3 to both sides. A0.5 for \(x > 5\).

The radius of the circle is half of its diameter: \(r = 12 \div 2 = 6\text{ cm}\). The area is calculated using the formula \(A = \pi r^2 = \pi \times 6^2 = 36\pi \approx 113.097\text{ cm}^2\). To 3 significant figures, this is \(113\text{ cm}^2\).

M1 for attempting to use \(\pi \times 6^2\) or \(\pi \times (12/2)^2\). A0.5 for \(113\) (accept 113.1 or 36\(\pi\)).

The sum of all possible probabilities must equal 1. Sum of given probabilities = \(0.15 + 0.35 + 0.20 = 0.70\). Probability of landing on D = \(1 - 0.70 = 0.30\).

M1 for \(1 - (0.15 + 0.35 + 0.20)\) or equivalent complete calculation. A0.5 for \(0.3\) or equivalent decimal/fraction.

Divide the numerical parts: \(3.8 \div 2 = 1.9\). Subtract the exponents of the power of 10 parts: \(10^5 \div 10^2 = 10^{5-2} = 10^3\). Combining these gives \(1.9 \times 10^3\).

M1 for an answer of the form \(1.9 \times 10^k\) where \(k \neq 3\), or converting to an ordinary number as \(1900\). A0.5 for \(1.9 \times 10^3\).

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

M1 for expanding at least one bracket correctly to get \(6x - 15\) or \(2x + 8\). A0.5 for \(8x - 7\).

Since the measurement is taken to the nearest \(0.1\text{ kg}\), the limit of accuracy is half of this value, which is \(0.05\text{ kg}\). The lower bound is the measured value minus this limit: \(4.5 - 0.05 = 4.45\text{ kg}\).

M1 for identifying the limit of accuracy as \(0.05\text{ kg}\) or for performing \(4.5 - 0.05\). A0.5 for \(4.45\).

First, find the multiplier for a 12% reduction: \( 1 - 0.12 = 0.88 \). Next, multiply the original price by this multiplier: \( 450 \times 0.88 = 396 \). Alternatively, find 12% of 450: \( 0.12 \times 450 = 54 \), then subtract this from the original price: \( 450 - 54 = 396 \).

M1 for a complete method to find the reduced price, e.g., \( 450 \times 0.88 \) or \( 450 - 0.12 \times 450 \). A0.5 for the correct final answer of 396.

Multiply the map distance by the scale factor: \( 8 \times 25\,000 = 200\,000 \text{ cm} \). Convert centimetres to metres by dividing by 100: \( 200\,000 \div 100 = 2000 \text{ m} \). Convert metres to kilometres by dividing by 1000: \( 2000 \div 1000 = 2 \text{ km} \).

M1 for showing the scale multiplication \( 8 \times 25\,000 \) or for converting units correctly (e.g., dividing by 100\,000). A0.5 for 2.

The measurement is given to the nearest 0.1 metres. The interval for the error is half of this value, which is \( 0.05 \text{ metres} \). Therefore, the upper bound is found by adding 0.05 to the measured value: \( 3.8 + 0.05 = 3.85 \text{ metres} \).

M1 for identifying the error interval boundary of 0.05 or showing \( 3.8 + 0.05 \). A0.5 for 3.85.

Add 8 to both sides of the inequality: \( 5x < 30 \). Divide both sides by 5: \( x < 6 \).

M1 for isolating the x term correctly, e.g., showing \( 5x < 30 \) or finding the critical value of 6. A0.5 for the correct inequality \( x < 6 \).

First, calculate the total sum of the four numbers: \( 4 \times 14 = 56 \). Next, find the sum of the three known numbers: \( 9 + 11 + 18 = 38 \). Finally, subtract the sum of the known numbers from the total sum: \( 56 - 38 = 18 \).

M1 for calculating the total sum of 56 or the sum of the three known numbers as 38. A0.5 for 18.

The first term \( a \) is 5, and the common difference \( d \) is 3. The formula for the nth term of an arithmetic sequence is \( a + (n - 1)d \). For the 30th term, substitute \( n = 30 \): \( 5 + (30 - 1) \times 3 = 5 + 29 \times 3 = 5 + 87 = 92 \).

M1 for expressing the nth term as \( 3n + 2 \) or for substituting correctly into \( a + (n-1)d \). A0.5 for 92.

Let the distance from Town A to Town B be \(d\) km. The time taken to travel from A to B is \(\frac{d}{18}\) hours, and the time taken to travel from B to A is \(\frac{d}{12}\) hours. The total distance for the round trip is \(2d\) km. The total time taken is \(\frac{d}{18} + \frac{d}{12} = \frac{2d + 3d}{36} = \frac{5d}{36}\) hours. The average speed for the entire journey is \(\frac{\text{Total Distance}}{\text{Total Time}} = \frac{2d}{\frac{5d}{36}} = \frac{72}{5} = 14.4\text{ km/h}\).

M1: For setting up expressions for time taken in both directions, e.g., \(\frac{d}{18}\) and \(\frac{d}{12}\) (or choosing a numerical distance like \(36\text{ km}\) and calculating times of \(2\) hours and \(3\) hours). M1: For calculating the total distance divided by total time, e.g., \(\frac{72}{5}\) or \(\frac{2d}{\frac{5d}{36}}\). A1: For 14.4.

Let \(H\) be the set of students who study History and \(G\) be the set of students who study Geography. The number of students who study at least one of these subjects is \(80 - 12 = 68\). Using the formula \(n(H \cup G) = n(H) + n(G) - n(H \cap G)\), we have \(68 = 45 + 40 - n(H \cap G)\), which gives \(n(H \cap G) = 85 - 68 = 17\) students who study both. The probability that a student studies Geography, given that they study History, is \(P(G | H) = \frac{n(H \cap G)}{n(H)} = \frac{17}{45}\).

M1: For finding the number of students who study History or Geography (or both), e.g., \(80 - 12 = 68\). M1: For finding the number of students who study both subjects, e.g., \(45 + 40 - 68 = 17\). A1: For \(\frac{17}{45}\) (must be in simplest form).

The volume of the cylinder is \(V = \pi r^2 h = \pi \times 3^2 \times 8 = 72\pi\text{ cm}^3\). Since the metal is recast into a sphere, the volume of the sphere is also \(72\pi\text{ cm}^3\). Using the volume of a sphere formula: \(\frac{4}{3} \pi R^3 = 72\pi\). Dividing both sides by \(\pi\) gives \(\frac{4}{3} R^3 = 72\). Solving for \(R^3\), we get \(R^3 = 72 \times \frac{3}{4} = 54\). Taking the cube root of both sides, \(R = \sqrt[3]{54} \approx 3.77976...\text{ cm}\). To 3 significant figures, the radius is \(3.78\text{ cm}\).

M1: For calculating the volume of the cylinder as \(72\pi\) (or approx. 226.2). M1: For setting up the equation \(\frac{4}{3}\pi R^3 = 72\pi\) (or their volume of the cylinder). M1: For rearranging to find \(R^3 = 54\) (or approx. 54). A1: For 3.78 (accept 3.78 only, do not accept 3.8).

The total time for the first 4 races is \(4 \times 14.2 = 56.8\) seconds. The total time for all 5 races is \(5 \times 13.9 = 69.5\) seconds. The time taken for the 5th race is the difference between these two totals: \(69.5 - 56.8 = 12.7\) seconds.

M1: For calculating the total time of the first 4 races: \(4 \times 14.2 = 56.8\). M1: For calculating the total time of the 5 races: \(5 \times 13.9 = 69.5\). A1: For 12.7.

The area of the rectangle is given by \(\text{length} \times \text{width} = (x + 5)(2x - 3) = 35\). Expanding the brackets: \(2x^2 - 3x + 10x - 15 = 35\), which simplifies to \(2x^2 + 7x - 15 = 35\). Subtracting 35 from both sides gives the quadratic equation: \(2x^2 + 7x - 50 = 0\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 2\), \(b = 7\), and \(c = -50\): \(x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-50)}}{2(2)} = \frac{-7 \pm \sqrt{49 + 400}}{4} = \frac{-7 \pm \sqrt{449}}{4}\). Since the dimensions must be positive, \(x\) must be positive: \(x = \frac{-7 + 21.1896}{4} \approx 3.55\) (to 2 d.p.).

M1: For expanding \((x + 5)(2x - 3)\) to get at least three correct terms in \(2x^2 + 7x - 15\). M1: For rearranging into the form \(2x^2 + 7x - 50 = 0\). M1: For applying the quadratic formula correctly to their quadratic equation. A1: For 3.55 (accept 3.55 only).

Let the original price of the coat be \(P\). A reduction of 15% means the price becomes \(0.85P\). A subsequent reduction of 10% on this price means the final price is \(0.85P \times 0.90 = 0.765P\). We are given that the final price is %61.20, so \(0.765P = 61.20\). Solving for \(P\): \(P = \frac{61.20}{0.765} = 80\). The original price was %80.

M1: For writing a correct expression for the combined multiplier, e.g., \(0.85 \times 0.90\) or \(0.765\). M1: For setting up the equation \(0.765P = 61.20\) or equivalent division \(\frac{61.20}{0.765}\). A1: For 80.

The gradient of \(L_1\) is \(m_1 = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2\). Since line \(L_2\) is perpendicular to \(L_1\), its gradient is \(m_2 = -\frac{1}{2} = -0.5\). Using the equation of a line \(y - y_1 = m(x - x_1)\) with the point \((3, -1)\): \(y - (-1) = -0.5(x - 3) \implies y + 1 = -0.5x + 1.5 \implies y = -0.5x + 0.5\).

M1: For finding the gradient of \(L_1\) as 2. M1: For finding the perpendicular gradient as -0.5 (or their negative reciprocal of \(m_1\)). A1: For the equation \(y = -0.5x + 0.5\) (or equivalent form in \(y = mx + c\), e.g., \(y = -\frac{1}{2}x + \frac{1}{2}\)).

Let the exterior angle of the regular polygon be \(x\) degrees. The interior angle is \(11x\) degrees. Since the interior and exterior angles on a straight line add up to \(180^\circ\), we have: \(x + 11x = 180 \implies 12x = 180 \implies x = 15^\circ\). The sum of the exterior angles of any polygon is \(360^\circ\). Therefore, the number of sides \(n\) is: \(n = \frac{360}{15} = 24\).

M1: For setting up the equation \(x + 11x = 180\) or equivalent. M1: For finding the exterior angle to be \(15^\circ\) (or the interior angle to be \(165^\circ\)). A1: For 24.

Let the initial value of the investment be \(V\). The multiplier for a \(12\%\) increase is \(1.12\). The multiplier for a \(5\%\) decrease is \(0.95\). After the first two years, the combined multiplier is \(1.12 \times 0.95 = 1.064\). Let the multiplier for the third year be \(m\), where \(m = 1 + \frac{x}{100}\). The overall multiplier for the three years is \(1.1438\). This gives the equation: \(1.064 \times m = 1.1438\). Solving for \(m\): \(m = \frac{1.1438}{1.064} = 1.075\). Since \(m = 1.075\), the percentage increase is \(7.5\%\), so \(x = 7.5\).

M1: For finding the combined multiplier for the first two years: \(1.12 \times 0.95 = 1.064\) (or equivalent using a trial value, e.g., \(100 \times 1.12 \times 0.95 = 106.4\)). M1: For setting up the equation to find the third year's multiplier: \(1.064 \times m = 1.1438\) or \(m = \frac{1.1438}{1.064}\). A1: \(7.5\) (accept \(7.5\%\)).

First, find the lengths of the two perpendicular sides of the triangular cross-section. Using trigonometry: \(AB = 12 \cos(35^\circ) \approx 9.8298\text{ cm}\) and \(BC = 12 \sin(35^\circ) \approx 6.8829\text{ cm}\). The area of the cross-section (triangle \(ABC\)) is \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 9.8298 \times 6.8829 \approx 33.829\text{ cm}^2\). The volume of the prism is the cross-sectional area multiplied by the length: \(\text{Volume} = 33.829 \times 15 \approx 507.43\text{ cm}^3\). To 3 significant figures, the volume is \(507\text{ cm}^3\).

M1: For using trigonometry to find at least one of the perpendicular sides: \(12 \cos(35^\circ)\) or \(12 \sin(35^\circ)\). M1: For finding both perpendicular sides: \(AB \approx 9.83\text{ cm}\) and \(BC \approx 6.88\text{ cm}\) (or equivalent correct method for the triangle's area, e.g., \(\frac{1}{2} \times 12 \cos(35^\circ) \times 12 \sin(35^\circ)\)). M1: For multiplying their cross-sectional area by \(15\). A1: \(507\) (accept answers in range \([507, 508]\)).

First, find the midpoints of the intervals: \(5\), \(15\), \(30\), and \(45\). Write an expression for the total estimated distance: \((8 \times 5) + (12 \times 15) + (w \times 30) + (10 \times 45) = 40 + 180 + 30w + 450 = 670 + 30w\). Write an expression for the total frequency: \(8 + 12 + w + 10 = 30 + w\). Set up the equation for the estimated mean: \(\frac{670 + 30w}{30 + w} = 25.4\). Multiply both sides by \(30 + w\): \(670 + 30w = 25.4(30 + w)\). Expand the brackets: \(670 + 30w = 762 + 25.4w\). Rearrange to solve for \(w\): \(30w - 25.4w = 762 - 670 \Rightarrow 4.6w = 92 \Rightarrow w = 20\).

M1: For identifying the correct midpoints: \(5, 15, 30, 45\) (at least three correct). M1: For writing a correct expression for total distance: \(670 + 30w\). M1: For setting up the equation \(\frac{670 + 30w}{30 + w} = 25.4\) and expanding to \(670 + 30w = 762 + 25.4w\) or equivalent. A1: \(20\).

First, convert mass to kilograms: \(280\text{ g} = 0.28\text{ kg}\). Next, convert volume to cubic metres: \(350\text{ cm}^3 = 350 \times 10^{-6}\text{ m}^3 = 0.00035\text{ m}^3\). Calculate density: \(\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{0.28}{0.00035} = 800\text{ kg/m}^3\). Alternatively, find the density in \(\text{g/cm}^3\): \(\frac{280}{350} = 0.8\text{ g/cm}^3\). Multiply by 1000 to convert to \(\text{kg/m}^3\): \(0.8 \times 1000 = 800\text{ kg/m}^3\).

B1 for the correct answer of 800.

First, find the arc length: \(\text{Arc length} = \frac{135}{360} \times 2 \times \pi \times 8 = 6\pi \approx 18.85\text{ cm}\). The perimeter includes the arc length plus two radii: \(\text{Perimeter} = 18.85 + 8 + 8 = 34.85\text{ cm}\). Rounding to 1 decimal place gives \(34.8\text{ cm}\).

B1 for the correct answer of \(34.8\text{ cm}\).

Subtract 5 from both sides: \(-2x < 6\). Divide both sides by \(-2\) and reverse the inequality sign: \(x > -3\).

B1 for the correct answer of \(x > -3\).

The total of the five numbers is \(12 \times 5 = 60\). The sum of the four given numbers is \(7 + 15 + 8 + 16 = 46\). The fifth number is \(60 - 46 = 14\).

B1 for the correct answer of 14.

The value of the car after 3 years is calculated using compound depreciation: \(\text{Value} = 18500 \times (1 - 0.12)^3 = 18500 \times 0.88^3 = 12607.23\). Rounding to the nearest pound gives \(\pounds 12,607\).

B1 for the correct answer of \(\pounds 12,607\).

We look for factors of the form \((3x + a)(x + b)\) where \(a \times b = -8\) and \(3b + a = -10\). Testing the options, \((3x + 2)(x - 4) = 3x^2 - 12x + 2x - 8 = 3x^2 - 10x - 8\).

B1 for the correct factorisation \((3x + 2)(x - 4)\).

To find the lower bound, subtract half of the unit of accuracy: \(8.4 - 0.05 = 8.35\text{ cm}\). To find the upper bound, add half of the unit of accuracy: \(8.4 + 0.05 = 8.45\text{ cm}\). Therefore, the error interval is \(8.35 \le L < 8.45\).

M1 for identifying the correct limits \(8.35\) and \(8.45\), A0.5 for the correct notation \(8.35 \le L < 8.45\) (accept equivalent inequalities with correct bounds).

To find the bottling rate, divide the total number of bottles filled by the total time taken: \(135 \div 4.5 = 30\) bottles per minute.

M1 for calculating \(135 \div 4.5\), A0.5 for \(30\).

First, find the actual profit: \(\text{\pounds}312 - \text{\pounds}240 = \text{\pounds}72\). Next, express this profit as a percentage of the original purchase price: \(\frac{72}{240} \times 100 = 30\%\).

M1 for finding the profit of \(\text{\pounds}72\) and writing it as a fraction of the cost price, i.e., \(\frac{72}{240}\), A0.5 for \(30\).

Divide both sides of the equation by 5: \(x - 3 = 2.5\). Then, add 3 to both sides: \(x = 5.5\). Alternatively, expand the bracket first: \(5x - 15 = 12.5\), which leads to \(5x = 27.5\) and hence \(x = 5.5\).

M1 for a correct first step: either dividing both sides by 5 to get \(x - 3 = 2.5\) or expanding to get \(5x - 15 = 12.5\), A0.5 for \(5.5\) (or equivalent fraction \(\frac{11}{2}\)).

First, find the total score of the first five students: \(5 \times 8 = 40\). Next, find the total score of all six students: \(6 \times 8.5 = 51\). Subtract the first total from the second total to find the sixth student's mark: \(51 - 40 = 11\).

M1 for calculating at least one total correctly (e.g., finding \(40\) or \(51\)), A0.5 for \(11\).

The exterior angle of a regular octagon (which has 8 sides) is \(360^\circ \div 8 = 45^\circ\). Since interior and exterior angles sum to \(180^\circ\), the interior angle is \(180^\circ - 45^\circ = 135^\circ\). Alternatively, sum of interior angles is \((8 - 2) \times 180^\circ = 1080^\circ\), and dividing this by 8 gives \(135^\circ\).

M1 for a complete method to find the interior angle, such as calculating the exterior angle \(360 \div 8\) or using the interior angle sum formula \((8-2) \times 180\), A0.5 for \(135\).

The probability of an event and its complement always sum to 1. Therefore, the probability that the bus does not arrive late is \(1 - 0.15 = 0.85\).

M1 for subtracting the late probability from 1: \(1 - 0.15\), A0.5 for \(0.85\) (accept equivalent fractions).

Substitute \(n = 5\) into the given expression: \(3(5)^2 - 5 = 3(25) - 5 = 75 - 5 = 70\).

M1 for substituting \(n=5\) into the formula, showing \(3 \times 5^2 - 5\), A0.5 for \(70\).

Let the original price be \(P\). The laptop is reduced by 12.5%, so the sale price is \(100\% - 12.5\% = 87.5\%\) of the original price. Thus, \(0.875 \times P = 364\). To find \(P\), we calculate \(P = \frac{364}{0.875} = 416\).

M1 for \(364 \div 0.875\) or equivalent. A0.5 for \(416\).

Since \(y\) is inversely proportional to the square of \(x\), we can write the formula as \(y = \frac{k}{x^2}\) where \(k\) is a constant. Substituting the given values: \(9 = \frac{k}{4^2} \implies 9 = \frac{k}{16} \implies k = 9 \times 16 = 144\). Now, substitute \(x = 6\) into the formula: \(y = \frac{144}{6^2} = \frac{144}{36} = 4\).

M1 for finding the constant of proportionality \(k = 144\) or writing \(y \times x^2 = 144\). A0.5 for \(4\).

Expanding the brackets first: \(10x - 12.4 = 15.6\). Adding 12.4 to both sides: \(10x = 28\). Dividing by 10: \(x = 2.8\).

M1 for expanding brackets correctly to \(10x - 12.4 = 15.6\) or dividing both sides by 4 to get \(2.5x - 3.1 = 3.9\). A0.5 for \(2.8\) (accept equivalent fractions).

The formula for the volume of a cylinder is \(V = \pi r^2 h\). Substitute the given values: \(V = \pi \times 4.5^2 \times 12 = \pi \times 20.25 \times 12 = 243\pi \approx 763.407\text{ cm}^3\). Rounding to 3 significant figures gives \(763\text{ cm}^3\).

M1 for substituting correctly into the volume formula: \(\pi \times 4.5^2 \times 12\). A0.5 for \(763\) (accept answers in range \(763\) to \(764\)).

The sum of all probabilities must equal 1. Thus, \(0.25 + 0.35 + 3x + x = 1\) which simplifies to \(0.60 + 4x = 1\). Solving this gives \(4x = 0.40\) so \(x = 0.1\). The probability of landing on C is \(3x = 3 \times 0.1 = 0.3\).

M1 for setting up the equation \(0.25 + 0.35 + 4x = 1\) or finding \(x = 0.1\). A0.5 for \(0.3\) (or equivalent fraction).

Since the measurement is given to the nearest millimetre (\(0.1\text{ cm}\)), the upper bound is found by adding half of this interval (\(0.05\text{ cm}\)) to the measurement. Upper bound = \(18.3 + 0.05 = 18.35\text{ cm}\).

M1 for identifying \(0.05\text{ cm}\) as the half-interval. A0.5 for \(18.35\).

Find the midpoints of the intervals: 5, 15, 25, and 35. Multiply each midpoint by its frequency: \(5 \times 4 = 20\), \(15 \times 8 = 120\), \(25 \times 5 = 125\), and \(35 \times 3 = 105\). The sum of these products is \(20 + 120 + 125 + 105 = 370\). Divide this by the total frequency of 20 to find the estimate of the mean: \(370 \div 20 = 18.5\).

M1 for finding midpoints and calculating \(\sum fx = 370\). A0.5 for \(18.5\).

The first differences are 7, 11, 15, 19. The second differences are 4, 4, 4. Since the second difference is 4, the coefficient of \(n^2\) is \(4 \div 2 = 2\). Subtracting \(2n^2\) from each term of the sequence gives the linear sequence: \(4 - 2(1)^2 = 2\), \(11 - 2(2)^2 = 3\), \(22 - 2(3)^2 = 4\), etc. The \(n\)-th term of this linear sequence is \(n + 1\). Combining these gives the formula \(2n^2 + n + 1\).

M1 for identifying the second difference is 4 and deducing the term \(2n^2\) or finding the linear sequence. A0.5 for \(2n^2 + n + 1\).

First, calculate the amount in Euros Liam receives: \(400 \times 1.15 = 460\) Euros. Next, subtract the amount spent to find the remaining Euros: \(460 - 160 = 300\) Euros. Finally, convert the remaining Euros back into pounds: \(300 \div 1.20 = 250\) pounds.

Method mark (1 mark): For correctly calculating the remaining Euros, showing \((400 \times 1.15) - 160 = 300\) Euros. Accuracy mark (0.5 marks): For dividing their remaining Euros by 1.20 to obtain the correct final answer of £250 (accept 250).

First, expand the squared bracket: \((3x - 5)^2 = 9x^2 - 30x + 25\). Next, expand the second term: \(-2x(x - 4) = -2x^2 + 8x\). Finally, combine like terms: \((9x^2 - 2x^2) + (-30x + 8x) + 25 = 7x^2 - 22x + 25\).

Method mark (1 mark): For expanding at least one part correctly to obtain either \(9x^2 - 30x + 25\) or \(-2x^2 + 8x\). Accuracy mark (0.5 marks): For combining terms correctly to give the final simplified expression \(7x^2 - 22x + 25\).

Find the total score of the first 5 students: \(5 \times 12 = 60\) marks. Find the total score of all 6 students: \(6 \times 13.5 = 81\) marks. Subtract the first total from the second total to find the 6th student's score: \(81 - 60 = 21\) marks.

Method mark (1 mark): For a correct method to calculate at least one of the total sums, i.e., \(5 \times 12 = 60\) or \(6 \times 13.5 = 81\). Accuracy mark (0.5 marks): For subtracting the two totals to find the correct score of 21.

First, calculate the arc length of the sector: Arc length = \(\frac{45}{360} \times 2 \times \pi \times 8 = 2\pi \approx 6.283\text{ cm}\). The total perimeter is the arc length plus the two straight edges (radii): Total perimeter = \(2\pi + 8 + 8 = 22.283...\text{ cm}\). Rounded to 2 decimal places, this is 22.28 cm.

Method mark (1 mark): For calculating the correct arc length of the sector, showing \(\frac{45}{360} \times 2 \times \pi \times 8\) (accept answers rounding to 6.28). Accuracy mark (0.5 marks): For adding 16 to their arc length and rounding correctly to 2 decimal places to get 22.28.

The sale price represents \(100\% - 15\% = 85\%\) of the original price. To find the original price, divide the sale price by 0.85: \(54.40 \div 0.85 = 64\). The original price was £64.

Method mark (1 mark): For setting up a correct division or equation, such as \(54.40 \div 0.85\) or equating \(85\% = 54.40\). Accuracy mark (0.5 marks): For the correct final price of £64 (accept 64).

First, expand the bracket on the right-hand side: \(4x - 7 < 2x + 10\). Next, subtract \(2x\) from both sides: \(2x - 7 < 10\). Add 7 to both sides: \(2x < 17\). Finally, divide by 2: \(x < 8.5\) (or \(x < \frac{17}{2}\)).

Method mark (1 mark): For expanding and correctly rearranging terms to obtain an inequality of the form \(2x < c\) (where \(c\) is a constant, e.g., \(2x < 17\)). Accuracy mark (0.5 marks): For the fully correct inequality \(x < 8.5\) or \(x < 17/2\).

First, convert all quantities to grams: 1.125 kg = 1125 g of flour. Next, calculate how many cookies can be made with each ingredient. Flour: \(1125 \div (250 \div 12) = 54\) cookies. Butter: \(700 \div (150 \div 12) = 56\) cookies. Sugar: \(500 \div (100 \div 12) = 60\) cookies. The limiting ingredient is flour, which allows for a maximum of 54 whole cookies.

M1: for converting 1.125 kg to 1125 g. M1: for finding the cookies limit for at least one ingredient. A1: for 54.

The volume of a cylinder is given by \(V = \pi r^2 h\). Substituting the given values: \(540\pi = \pi r^2 \times 15\). Dividing both sides by \(15\pi\) gives \(r^2 = 36\), which means the radius \(r = 6\text{ cm}\). The total surface area of a cylinder is \(A = 2\pi r^2 + 2\pi r h\). Substituting \(r = 6\) and \(h = 15\): \(A = 2\pi(6)^2 + 2\pi(6)(15) = 72\pi + 180\pi = 252\pi\text{ cm}^2\).

M1: for setting up the volume equation and finding \(r = 6\). M1: for substituting \(r = 6\) and \(h = 15\) into the surface area formula. A1: for \(252\pi\).

Multiply the first equation by 3 to get \(9x + 6y = 21\). Multiply the second equation by 2 to get \(8x - 6y = 30\). Adding these equations yields \(17x = 51\), so \(x = 3\). Substitute \(x = 3\) into the first equation: \(3(3) + 2y = 7 \implies 9 + 2y = 7 \implies 2y = -2 \implies y = -1\). Therefore, \(x + y = 3 + (-1) = 2\).

M1: for a correct method to eliminate one variable. M1: for finding the correct value of either \(x = 3\) or \(y = -1\). A1: for 2.

The total probability of all outcomes must sum to 1. The probability of picking either a blue or yellow counter is \(1 - 0.4 = 0.6\). Let the probability of choosing a yellow counter be \(p\). Since there are twice as many blue counters as yellow, the probability of choosing a blue counter is \(2p\). This gives the equation \(2p + p = 0.6 \implies 3p = 0.6 \implies p = 0.2\).

M1: for finding the combined probability of blue and yellow is 0.6. M1: for setting up the ratio equation \(3p = 0.6\). A1: for 0.2.

The sale price represents 85% of the original price, so the original price is \(578 \div 0.85 = 680\). The final price is a 10% increase from the sale price, so the final price is \(578 \times 1.10 = 635.80\). The difference between the original price and the final price is \(680 - 635.80 = 44.20\).

M1: for calculating the original price as 680. M1: for calculating the final price as 635.80. A1: for 44.20.

The midpoints for the four intervals are 5, 15, 25, and 35. Multiply each midpoint by its frequency: \(5 \times 6 = 30\), \(15 \times 14 = 210\), \(25 \times 15 = 375\), and \(35 \times 5 = 175\). The sum of these products is \(30 + 210 + 375 + 175 = 790\). The estimate of the mean is the sum of the products divided by the total frequency: \(790 \div 40 = 19.75\) minutes.

M1: for identifying the correct midpoints. M1: for calculating the sum of \(\text{midpoint} \times \text{frequency}\). A1: for 19.75.

First, multiply both sides by \(x - 2\): \(y(x - 2) = 3x + 5\). Expand the left side: \(xy - 2y = 3x + 5\). Rearrange the terms to get all terms with \(x\) on one side: \(xy - 3x = 2y + 5\). Factorise \(x\) from the left side: \(x(y - 3) = 2y + 5\). Finally, divide by \(y - 3\) to get \(x = \frac{2y + 5}{y - 3}\).

M1: for multiplying by \(x-2\) and expanding. M1: for isolating the \(x\) terms on one side and factorising. A1: for \(x = \frac{2y + 5}{y - 3}\) or equivalent.

Find the 10th term by substituting \(n = 10\): \(3(10)^2 - 5(10) + 2 = 300 - 50 + 2 = 252\). Find the 8th term by substituting \(n = 8\): \(3(8)^2 - 5(8) + 2 = 3(64) - 40 + 2 = 192 - 40 + 2 = 154\). Calculate the difference: \(252 - 154 = 98\).

M1: for substituting \(n = 10\) or \(n = 8\) into the formula. M1: for correctly finding 252 and 154. A1: for 98.

Let the volume of copper be \(3x\text{ cm}^3\) and the volume of zinc be \(5x\text{ cm}^3\).

The total volume of the prism is:
\[3x + 5x = 8x\text{ cm}^3\]

Using the formula \(\text{Mass} = \text{Density} \times \text{Volume}\):
\[\text{Mass of copper} = 8.9 \times 3x = 26.7x\text{ g}\]
\[\text{Mass of zinc} = 7.1 \times 5x = 35.5x\text{ g}\]

Therefore, the total mass of the prism is:
\[26.7x + 35.5x = 62.2x\text{ g}\]

Now, calculate the average density of the prism:
\[\text{Average density} = \frac{\text{Total mass}}{\text{Total volume}} = \frac{62.2x}{8x} = 7.775\text{ g/cm}^3\]

Rounding to 2 decimal places gives \(7.78\text{ g/cm}^3\).

M1: For a correct method to find the total mass for any chosen volume in the ratio \(3:5\), e.g. \(3 \times 8.9 + 5 \times 7.1\) or showing \(26.7\) and \(35.5\).
M1: For dividing their total mass by their total volume, e.g. \(\frac{62.2}{8}\).
A1: For \(7.78\) (accept \(7.775\)).

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

For Sector A:
\[\text{Area}_A = \frac{x}{360} \times \pi r^2 = 40\]

For Sector B:
\[\text{Area}_B = \frac{0.6x}{360} \times \pi (1.5r)^2\]

Expand and simplify the expression for Sector B:
\[\text{Area}_B = \frac{0.6x}{360} \times \pi \times 2.25r^2\]
\[\text{Area}_B = 0.6 \times 2.25 \times \left( \frac{x}{360} \times \pi r^2 \right)\]
\[\text{Area}_B = 1.35 \times \left( \frac{x}{360} \times \pi r^2 \right)\]

Substitute the known area of Sector A into the equation:
\[\text{Area}_B = 1.35 \times 40 = 54\text{ cm}^2\]

M1: For writing a correct expression for the area of both sectors, e.g. \(\frac{x}{360} \pi r^2 = 40\) and \(\frac{0.6x}{360} \pi (1.5r)^2\).
M1: For simplifying the expression of Sector B to find its ratio to Sector A, e.g. \(0.6 \times 2.25 \times 40\) or \(1.35 \times 40\).
A1: For \(54\).

The probability of choosing a white sock first is \(\frac{6}{n}\).
Since the first sock is not replaced, the probability of choosing a second white sock is \(\frac{5}{n-1}\).

The probability that both socks are white is:
\[\frac{6}{n} \times \frac{5}{n-1} = \frac{30}{n(n-1)}\]

We are given that this probability is \(\frac{1}{3}\):
\[\frac{30}{n(n-1)} = \frac{1}{3}\]

Cross-multiplying gives:
\[n(n-1) = 90\]
\[n^2 - n - 90 = 0\]

Factorising the quadratic equation:
\[(n - 10)(n + 9) = 0\]

Since the number of socks \(n\) must be positive, we have:
\[n = 10\]

There are 10 socks in total. Since 6 of them are white, the number of black socks is:
\[10 - 6 = 4\]

M1: For writing a correct algebraic expression for the probability of selecting two white socks, e.g. \(\frac{6}{n} \times \frac{5}{n-1}\).
M1: For setting up the equation \(\frac{30}{n(n-1)} = \frac{1}{3}\) and expanding to form a quadratic equation, e.g. \(n^2 - n - 90 = 0\).
M1: For solving the quadratic equation to find \(n = 10\).
A1: For \(4\).

To find the original price, we divide the sale price by the multiplier for a 10% decrease, which is \(0.9\). \(\text{Original Price} = 468 \div 0.9 = 520\). The original price was £520.

B1: for correct answer £520

Expand the brackets first: \(5x - 3 > 3x + 15\). Subtract \(3x\) from both sides to get \(2x - 3 > 15\). Add 3 to both sides to get \(2x > 18\). Divide by 2 to get \(x > 9\).

B1: for correct inequality \(x > 9\)

Since \(y \propto \frac{1}{x^2}\), we can write \(y = \frac{k}{x^2}\) for some constant \(k\). Substitute \(x = 3\) and \(y = 8\): \(8 = \frac{k}{3^2} \implies k = 72\). Now substitute \(k = 72\) and \(x = 6\): \(y = \frac{72}{6^2} = \frac{72}{36} = 2\).

B1: for correct answer 2

The volume \(V\) of a cylinder is given by \(V = \pi r^2 h\). Substituting the given values: \(V = \pi \times 4^2 \times 9 = 144\pi \approx 452.389\text{ cm}^3\). To 3 significant figures, this is \(452\text{ cm}^3\).

B1: for correct answer \(452\text{ cm}^3\)

With 5 numbers, when arranged in ascending order, the median is the 3rd number. If \(x = 17\), the ordered list is \(12\), \(15\), \(17\), \(18\), \(22\) and the 3rd number (median) is indeed 17. For other options: if \(x = 14\) median is 15; if \(x = 16\) median is 16; if \(x = 19\) median is 18.

B1: for correct option 17

The probability that a seed does not germinate is \(1 - 0.8 = 0.2\). For exactly one to germinate, the possible mutually exclusive outcomes are: first germinates and second does not (\(0.8 \times 0.2 = 0.16\)) or first does not and second germinates (\(0.2 \times 0.8 = 0.16\)). The total probability is \(0.16 + 0.16 = 0.32\).

B1: for correct answer 0.32

The gradient \(m\) is given by \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Substituting the coordinates: \(m = \frac{17 - 5}{4 - (-2)} = \frac{12}{6} = 2\).

B1: for correct gradient 2