2022 AQA GCSE Mathematics 8300 模擬試題連答案詳解

Using the order of operations (BIDMAS), we calculate indices first, then multiplication, and finally subtraction. \(3^3 = 27\), \(4^2 = 16\). The expression becomes \(27 - 2 \times 16 = 27 - 32 = -5\).

B1: For the correct answer of \(-5\) only.

List the prime factors of both numbers: \(140 = 2 \times 2 \times 5 \times 7\) and \(210 = 2 \times 3 \times 5 \times 7\). The common prime factors are 2, 5, and 7. The HCF is found by multiplying these common prime factors together: \(2 \times 5 \times 7 = 70\).

M1 for listing prime factors of at least one number correctly or listing common factors. A1 for 70.

The total number of parts is \(1 + 3 + 5 = 9\). The value of one part is \(£180 \div 9 = £20\). The smallest share is \(1 \times £20 = £20\). The largest share is \(5 \times £20 = £100\). The difference between them is \(£100 - £20 = £80\).

M1 for \(180 \div 9\) (or 20) or identifying the individual shares as £20 and £100. A1 for £80.

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

M1 for at least one correct bracket expansion: \(6x - 15\) or \(-2x + 8\). A1 for \(4x - 7\).

Convert both mixed numbers to improper fractions: \(2\frac{1}{3} = \frac{7}{3}\) and \(1\frac{3}{4} = \frac{7}{4}\). Divide the fractions by multiplying by the reciprocal: \(\frac{7}{3} \times \frac{4}{7} = \frac{28}{21} = \frac{4}{3}\). Convert the improper fraction back to a mixed number: \(1\frac{1}{3}\).

M1 for converting both mixed numbers to improper fractions: \(\frac{7}{3}\) and \(\frac{7}{4}\). A1 for \(1\frac{1}{3}\) (accept \(1 \frac{1}{3}\)).

The total of all five scores is \(4 + 5 + 5 + 7 + x = 21 + x\). Since there are 5 students and their mean score is 6, the sum of all scores must be \(5 \times 6 = 30\). Set up the equation: \(21 + x = 30\). Solving this gives \(x = 9\).

M1 for finding that the total score is 30, or writing the equation \(21 + x = 30\). A1 for 9.

First expand the bracket on the right side: \(5y - 3 > 2y + 12\). Subtract \(2y\) from both sides: \(3y - 3 > 12\). Add 3 to both sides: \(3y > 15\). Divide both sides by 3 to get \(y > 5\).

M1 for expanding the bracket to get \(2y + 12\) and attempting to collect terms. A1 for \(y > 5\).

The total probability of all mutually exclusive events must sum to 1. The sum of the probabilities of choosing red and blue is \(0.35 + 0.4 = 0.75\). The probability of choosing yellow is \(1 - 0.75 = 0.25\).

M1 for \(0.35 + 0.4\) or \(1 - (0.35 + 0.4)\). A1 for 0.25 (accept equivalent fractions or percentages, such as \(\frac{1}{4}\) or 25%).

Use the formula for the area of a trapezium: \(\text{Area} = \frac{1}{2}(a + b)h\). Substituting the given values: \(\text{Area} = \frac{1}{2}(6 + 10) \times 5 = \frac{1}{2}(16) \times 5 = 8 \times 5 = 40\text{ cm}^2\).

M1 for substituting the values correctly into the trapezium formula: \(\frac{6 + 10}{2} \times 5\). A1 for 40.

First, evaluate the numerator: \(4^3 - 2^4 = 64 - 16 = 48\). Next, evaluate the expression inside the square root in the denominator: \(12^2 - 80 = 144 - 80 = 64\). Now find the square root of this value: \(\sqrt{64} = 8\). Finally, divide the numerator by the denominator: \(\frac{48}{8} = 6\).

M1 for correctly evaluating the numerator to 48 OR the denominator to 8
A1 for 6

First, expand the bracket on the left side of the inequality: \(5x - 10 \le 2x + 8\). Next, subtract \(2x\) from both sides: \(3x - 10 \le 8\). Then, add \(10\) to both sides: \(3x \le 18\). Finally, divide both sides by \(3\) to find the solution: \(x \le 6\).

M1 for correctly expanding the bracket to \(5x - 10\) OR for isolating the x terms on one side to get \(3x \le 18\)
A1 for \(x \le 6\) (accept equivalent notation like \(x \le 6\) or \(x \leq 6\))

The ratio of red to yellow counters is \(3 : 5\). The difference in ratio parts between yellow and red counters is \(5 - 3 = 2\) parts. We are given that this difference represents 24 counters, so 2 parts = 24 counters. This means 1 part = 12 counters. The number of blue counters is represented by 4 parts in the ratio, so the number of blue counters is \(4 \times 12 = 48\).

M1 for finding that the difference of 2 parts corresponds to 24 counters (e.g. showing \(24 \div 2\) or \(12\))
A1 for 48

First, find the highest common numerical factor of 12 and 18, which is 6. Next, find the highest common algebraic factors of \(a^2b\) and \(ab^2\), which is \(ab\). Combined, the highest common factor is \(6ab\). Dividing each term by \(6ab\), we get \(12a^2b \div 6ab = 2a\) and \(-18ab^2 \div 6ab = -3b\). Thus, the fully factorised expression is \(6ab(2a - 3b)\).

M1 for identifying a common factor of at least \(6\), \(ab\), \(6a\), or \(6b\), resulting in a partially factorised expression such as \(6(2a^2b - 3ab^2)\) or \(ab(12a - 18b)\)
A1 for \(6ab(2a - 3b)\)

First, convert both mixed numbers into improper fractions: \(2\frac{2}{3} = \frac{8}{3}\) and \(1\frac{4}{5} = \frac{9}{5}\). Next, multiply the improper fractions: \(\frac{8}{3} \times \frac{9}{5} = \frac{72}{15}\). Simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, 3, to get \(\frac{24}{5}\). Finally, convert \(\frac{24}{5}\) back into a mixed number: \(24 \div 5 = 4\) with a remainder of 4, which gives \(4\frac{4}{5}\).

M1 for converting both mixed numbers to correct improper fractions (\(\frac{8}{3}\) and \(\frac{9}{5}\)) and showing intention to multiply
A1 for \(4\frac{4}{5}\) (accept equivalent mixed fraction in simplest form)

The sum of all probabilities for the spinner must equal 1. First, find the probability of landing on 4: \(1 - (0.25 + 0.4 + 0.15) = 1 - 0.8 = 0.2\). Next, to find the estimated number of times it lands on 4 in 200 spins, multiply the probability by the number of spins: \(200 \times 0.2 = 40\).

M1 for finding the probability of landing on 4 is \(0.2\) OR for showing a method to multiply their probability by 200
A1 for 40

To keep the ratio of juice to water at \(3 : 5\), we find the limiting ingredient. If we use all \(12\) litres of juice, we would need \(12 \times \frac{5}{3} = 20\) litres of water. But we only have \(18\) litres of water. Therefore, water is the limiting ingredient and we must use all \(18\) litres of water. For \(18\) litres of water, the volume of juice needed is \(18 \times \frac{3}{5} = 10.8\) litres. The maximum volume of the drink is \(18 + 10.8 = 28.8\) litres.

M1 for finding the required amount of water for 12 litres of juice (\(20\) litres) or the required amount of juice for 18 litres of water (\(10.8\) litres). M1 for adding 18 litres of water to their calculated juice amount (e.g. \(18 + 10.8\)). A1 for \(28.8\) (accept \(28\frac{4}{5}\)).

First, find the fraction of students who play a string instrument: \(\frac{2}{5} \times \frac{3}{8} = \frac{6}{40} = \frac{3}{20}\). Next, subtract this fraction from 1 to find those who do not play a string instrument: \(1 - \frac{3}{20} = \frac{17}{20}\).

M1 for multiplying \(\frac{2}{5} \times \frac{3}{8}\). A1 for \(\frac{3}{20}\) (or equivalent fraction). M1 for subtracting their fraction from 1 (e.g., \(1 - \frac{3}{20}\)). A1 for \(\frac{17}{20}\) (must be in simplest form).

First, calculate the arc length of the sector: \(\text{Arc length} = \frac{\theta}{360} \times 2\pi r = \frac{120}{360} \times 2 \times \pi \times 6 = \frac{1}{3} \times 12\pi = 4\pi\text{ cm}\). The perimeter of the sector includes the arc length and two radii: \(\text{Perimeter} = \text{Arc length} + 2r = 4\pi + 2(6) = 12 + 4\pi\text{ cm}\).

M1 for \(\frac{120}{360} \times 2 \times \pi \times 6\) or showing arc length is \(4\pi\). M1 for adding \(2 \times 6\) to their arc length. A1 for \(12 + 4\pi\) (or \(4\pi + 12\)).

Multiply the first equation by 5 and the second by 2 to equate the y coefficients: \(15x - 10y = 80\) and \(4x + 10y = -4\). Add these two equations together: \(19x = 76\), which simplifies to \(x = 4\). Substitute \(x = 4\) into the first equation: \(3(4) - 2y = 16 \implies 12 - 2y = 16 \implies -2y = 4 \implies y = -2\). The solution is \(x = 4\) and \(y = -2\).

M1 for a correct process to eliminate one variable (e.g. multiplying equations to get coefficients equal and adding/subtracting). A1 for \(x = 4\) or \(y = -2\). A1 for both \(x = 4\) and \(y = -2\).

We need two numbers that multiply to \(3 \times (-5) = -15\) and add to \(-14\). These numbers are \(-15\) and \(1\). Split the middle term: \(3x^2 - 15x + x - 5\). Factorise by grouping: \(3x(x - 5) + 1(x - 5) = (3x + 1)(x - 5)\).

M1 for splitting the middle term correctly (e.g. \(-15x + x\)) or showing two brackets of the form \((3x + a)(x + b)\) where \(ab = -5\) or \(a + 3b = -14\). M1 for partial factorisation such as \(3x(x - 5) + 1(x - 5)\). A1 for \((3x + 1)(x - 5)\) (or \((x - 5)(3x + 1)\)).

The total probability of all outcomes is 1. The probability of choosing either a blue or a green counter is \(1 - 0.3 = 0.7\). This remaining probability is shared between blue and green in the ratio \(3 : 4\). Total parts = \(3 + 4 = 7\). Value of one part = \(0.7 \div 7 = 0.1\). Probability of green = \(4 \times 0.1 = 0.4\).

M1 for calculating the combined probability of blue and green as \(0.7\) (or \(1 - 0.3\)). M1 for dividing their combined probability by \(7\) and multiplying by \(4\). A1 for \(0.4\) (accept equivalent fractions or percentages, e.g. \(\frac{2}{5}\) or \(40\%\)).

Find the first differences: \(7, 11, 15, 19\). Find the second differences: \(4, 4, 4\). Since the second difference is constant at \(4\), the coefficient of \(n^2\) is \(4 \div 2 = 2\). Subtract \(2n^2\) from the original sequence: For \(n = 1: 4 - 2(1)^2 = 2\); For \(n = 2: 11 - 2(2)^2 = 3\); For \(n = 3: 22 - 2(3)^2 = 4\); For \(n = 4: 37 - 2(4)^2 = 5\). The linear sequence is \(2, 3, 4, 5, \dots\), which has the \(n\)-th term \(n + 1\). Combining these gives the final \(n\)-th term: \(2n^2 + n + 1\).

M1 for finding the second difference of \(4\) and identifying the \(2n^2\) term. M1 for setting up a linear sequence by subtracting \(2n^2\) from the terms (getting \(2, 3, 4, 5, \dots\)) and attempting to find its \(n\)-th term. A1 for \(2n^2 + n + 1\) (or equivalent).

Find the total height of the girls: \(8 \times 155 = 1240\text{ cm}\). Find the total height of the boys: \(12 \times 165 = 1980\text{ cm}\). Find the total combined height: \(1240 + 1980 = 3220\text{ cm}\). Find the combined mean height for all 20 children: \(3220 \div 20 = 161\text{ cm}\).

M1 for calculating the total height of either group (i.e., \(1240\) or \(1980\)). M1 for dividing the total combined height by \(20\) (e.g. \(\frac{1240 + 1980}{20}\)). A1 for \(161\text{ cm}\).

1. Find the remaining fraction of books after subtracting the fiction books:
\(1 - \frac{3}{8} = \frac{5}{8}\)

2. Calculate the fraction of non-fiction books:
40% of \(\frac{5}{8}\) is:
\(\frac{40}{100} \times \frac{5}{8} = \frac{2}{5} \times \frac{5}{8} = \frac{2}{8} = \frac{1}{4}\)

3. Subtract non-fiction from the remaining fraction to find the children's books:
\(\frac{5}{8} - \frac{2}{8} = \frac{3}{8}\)

M1: For finding the remaining fraction \(\frac{5}{8}\) or writing a correct expression.
M1: For multiplying 40% (or equivalent fraction) by their remaining fraction.
A1: For \(\frac{3}{8}\) or equivalent fully simplified fraction.

1. Write down the two ratios:
\(R : G = 3 : 5\)
\(G : Y = 2 : 3\)

2. Find a common value for the green sweets (G). The lowest common multiple of 5 and 2 is 10.

Multiply the first ratio by 2:
\(R : G = 6 : 10\)

Multiply the second ratio by 5:
\(G : Y = 10 : 15\)

3. Combine the ratios:
\(R : G : Y = 6 : 10 : 15\)

4. Find the ratio of red to yellow sweets:
\(R : Y = 6 : 15\)

Simplify this by dividing both parts by 3:
\(R : Y = 2 : 5\)

M1: For a correct method to find a common value for green sweets (e.g. scaling to G = 10).
M1: For combining to find the unsimplified ratio of red to yellow, \(6 : 15\).
A1: For the fully simplified ratio \(2 : 5\).

1. Factorise the numerator (difference of two squares):
\(x^2 - 9 = (x - 3)(x + 3)\)

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

3. Cancel the common factor \((x + 3)\):
\(\frac{(x - 3)(x + 3)}{(2x - 1)(x + 3)} = \frac{x - 3}{2x - 1}\)

M1: For factorising the numerator to \((x - 3)(x + 3)\).
M1: For factorising the denominator to \((2x - 1)(x + 3)\).
A1: For the fully simplified fraction \(\frac{x - 3}{2x - 1}\) or equivalent.

1. Calculate the volume of the cylinder:
\(V = \pi r_c^2 h = \pi \times 2^2 \times 18 = 72\pi\)

2. Set up the equation for the volume of the sphere and equate it:
\(\frac{4}{3}\pi r^3 = 72\pi\)

3. Solve for \(r\):
\(\frac{4}{3}r^3 = 72 \Rightarrow 4r^3 = 216 \Rightarrow r^3 = 54\)
\(r = \sqrt[3]{54}\)

M1: For finding the volume of the cylinder as \(72\pi\).
M1: For equating the volumes and solving for \(r^3\) (e.g. \(r^3 = 54\)).
A1: For the final answer \(\sqrt[3]{54}\).

1. Multiply the second equation by 2:
\(8x - 2y = 18\)

2. Add the two equations together to eliminate y:
\((3x + 2y) + (8x - 2y) = 4 + 18\)
\(11x = 22 \Rightarrow x = 2\)

3. Substitute \(x = 2\) back into the second equation:
\(4(2) - y = 9 \Rightarrow 8 - y = 9 \Rightarrow y = -1\)

M1: For a correct method to eliminate one variable (e.g. multiplying the second equation by 2 to align y coefficients).
M1: For finding the correct value of one variable (\(x = 2\) or \(y = -1\)).
A1: For both correct values: \(x = 2\) and \(y = -1\).

1. Set up simultaneous equations by substituting \(n = 2\) and \(n = 4\):
For \(n = 2\): \(a(2)^2 + b(2) = 10 \Rightarrow 4a + 2b = 10\) (or \(2a + b = 5\))
For \(n = 4\): \(a(4)^2 + b(4) = 36 \Rightarrow 16a + 4b = 36\) (or \(4a + b = 9\))

2. Solve the equations:
Subtract \(2a + b = 5\) from \(4a + b = 9\):
\(2a = 4 \Rightarrow a = 2\)

3. Find \(b\):
\(2(2) + b = 5 \Rightarrow 4 + b = 5 \Rightarrow b = 1\)

M1: For setting up at least one correct equation in terms of \(a\) and \(b\) (e.g., \(4a+2b=10\)).
M1: For a correct method to solve the simultaneous equations to find either \(a\) or \(b\).
A1: For both correct values: \(a = 2\) and \(b = 1\).

1. Find the total number of counters: \(3 + 7 = 10\).

2. Work out the probability of Red then Blue (RB):
\(P(RB) = \frac{3}{10} \times \frac{7}{9} = \frac{21}{90}\)

3. Work out the probability of Blue then Red (BR):
\(P(BR) = \frac{7}{10} \times \frac{3}{9} = \frac{21}{90}\)

4. Add the two probabilities together:
\(P(\text{different}) = \frac{21}{90} + \frac{21}{90} = \frac{42}{90}\)

5. Simplify the fraction:
\(\frac{42}{90} = \frac{7}{15}\)

M1: For a correct product of probabilities for one combination (e.g., \(\frac{3}{10} \times \frac{7}{9}\)).
M1: For adding both combinations: \(\frac{21}{90} + \frac{21}{90}\).
A1: For the simplified answer \(\frac{7}{15}\) (or equivalent).

Using the multiplier \(0.92\) for an \(8\%\) annual depreciation: After 8 years: \(15000 \times 0.92^8 \approx 7698.35\). After 9 years: \(15000 \times 0.92^9 \approx 7082.48\). Therefore, it takes 9 whole years for the value to fall below £7500.

M1 for attempting trial and improvement with \(0.92^n\) or writing down values for consecutive years. A1 for 9 years (C).

Factorise the numerator: \(3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2)\). Factorise the denominator: \(x^2 + 5x + 6 = (x + 2)(x + 3)\). Divide numerator and denominator by the common factor \((x + 2)\) to get \(\frac{3(x - 2)}{x + 3}\).

M1 for factorising either the numerator or denominator correctly. A1 for the correct simplified fraction (A).

Probability of Red then Red: \(P(\text{R, R}) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90}\). Probability of Blue then Blue: \(P(\text{B, B}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90}\). Probability of same colour: \(\frac{30}{90} + \frac{12}{90} = \frac{42}{90} = \frac{7}{15}\).

M1 for calculating either \(P(\text{R, R})\) or \(P(\text{B, B})\) without replacement, or for summing two correct products. A1 for \(\frac{7}{15}\) (A).

For the first rod, the upper bound of length \(L_1\) is \(12.45\text{ cm}\). For the second rod, the lower bound of length \(L_2\) is \(8.645\text{ cm}\). The upper bound for the difference is \(L_1(\text{upper}) - L_2(\text{lower}) = 12.45 - 8.645 = 3.805\text{ cm}\).

M1 for finding the upper bound of the first rod (\(12.45\)) or the lower bound of the second rod (\(8.645\)). A1 for 3.805 cm (C).

First, find the perpendicular height \(h\) using Pythagoras' theorem: \(h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}\). Use the cone volume formula: \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \pi \times 5^2 \times 12 = 100\pi \approx 314.159\text{ cm}^3\). To 3 significant figures, the volume is 314 cm³.

M1 for using Pythagoras' theorem to find the height is \(12\text{ cm}\). A1 for 314 cm³ (A).

Rearranging the equation of Line L into the form \(y = mx + c\): \(2y = 3x + 8 \implies y = \frac{3}{2}x + 4\). The gradient of Line L is \(\frac{3}{2}\). The perpendicular gradient must be \(-\frac{2}{3}\). Let's check the given options: option C is \(3y + 2x = 12 \implies 3y = -2x + 12 \implies y = -\frac{2}{3}x + 4\), which has a gradient of \(-\frac{2}{3}\).

M1 for finding the gradient of Line L is \(\frac{3}{2}\) or identifying that the perpendicular gradient is \(-\frac{2}{3}\). A1 for identifying the correct equation (C).

Find the first differences: 7, 11, 15. Find the second differences: 4, 4. The coefficient of \(n^2\) is half of the second difference: \(4 \div 2 = 2\). Subtract \(2n^2\) from the sequence terms: for \(n=1\), \(3 - 2(1)^2 = 1\); for \(n=2\), \(10 - 2(2)^2 = 2\); for \(n=3\), \(21 - 2(3)^2 = 3\). This leaves the sequence \(1, 2, 3, ...\), which is represented by \(n\). Thus, the \(n\)-th term is \(2n^2 + n\).

M1 for finding the second differences are 4 or identifying the \(2n^2\) term. A1 for the correct formula (A).

The total frequency is 40. The median is the average of the 20th and 21st values. The cumulative frequencies are: up to 10 mins: 5, up to 20 mins: 17, up to 30 mins: 32. Since 17 values are below 20 and 32 values are below 30, the 20th and 21st values lie in the interval \(20 < t \le 30\).

M1 for calculating cumulative frequencies or identifying the position of the median as the 20th/21st value. A1 for identifying the correct interval (B).

Since \(y\) is inversely proportional to the square of \(x\), we can write the equation as \(y = \frac{k}{x^2}\) where \(k\) is a constant. Substituting \(x = 4\) and \(y = 7.5\) gives \(7.5 = \frac{k}{4^2}\), which simplifies to \(7.5 = \frac{k}{16}\). Multiplying both sides by 16 gives \(k = 7.5 \times 16 = 120\). Therefore, the formula is \(y = \frac{120}{x^2}\). When \(x = 5\), we have \(y = \frac{120}{5^2} = \frac{120}{25} = 4.8\).

M1 for setting up the correct relationship equation \(y = \frac{k}{x^2}\) or finding \(k = 120\). A1 for the correct answer of 4.8.

The total surface area of a solid cylinder is given by the formula \(A = 2\pi r^2 + 2\pi r h\). Substituting \(r = 3.5\) and \(h = 8.2\), we get \(A = 2\pi (3.5)^2 + 2\pi (3.5)(8.2)\). This simplifies to \(A = 24.5\pi + 57.4\pi = 81.9\pi \approx 257.296\text{ cm}^2\). To 3 significant figures, this is \(257\text{ cm}^2\).

M1 for substituting values into the total surface area formula \(2\pi r^2 + 2\pi rh\). A1 for 257.

The number of students who study History is 22. Since 6 students study both subjects, the number of students who study History but not Geography is \(22 - 6 = 16\). The probability that a randomly chosen student from the group of 40 studies History but not Geography is \(\frac{16}{40} = 0.4\).

M1 for subtracting 6 from 22 to find the number of students studying History only (16). A1 for the correct probability of 0.4.

First, calculate the numerator: \(\sqrt{8.5^2 + 19.2^2} = \sqrt{72.25 + 368.64} = \sqrt{440.89} = 21\). Next, calculate the denominator: \(3.1 \times 1.4 = 4.34\). Divide the numerator by the denominator: \(21 \div 4.34 \approx 4.8387\). Rounding to 3 significant figures gives 4.84.

M1 for finding 21 or 4.34. A1 for 4.84 (accept 4.839).

The sale price of £442 represents 85% of the original price because 100% - 15% = 85%. To find the original price, divide £442 by 0.85: \(442 \div 0.85 = 520\).

M1 for \(442 \div 0.85\) or \(442 \div 85 \times 100\). A1 for 520.

Expand the first bracket: \(5 \times 2x - 5 \times 3 = 10x - 15\). Expand the second bracket: \(-3 \times x - 3 \times (-4) = -3x + 12\). Combine like terms: \((10x - 3x) + (-15 + 12) = 7x - 3\).

M1 for expanding to get \(10x - 15\) or \(-3x + 12\) (allow one sign error). A1 for \(7x - 3\) or equivalent.

Expand the bracket: \(12x - 20 = 22\). Add 20 to both sides: \(12x = 42\). Divide by 12: \(x = \frac{42}{12} = 3.5\). Alternatively, divide by 4 first: \(3x - 5 = 5.5\), then add 5: \(3x = 10.5\), and divide by 3: \(x = 3.5\).

M1 for \(12x - 20 = 22\) or \(3x - 5 = 5.5\). A1 for 3.5 (or \(\frac{7}{2}\) or equivalent fraction).

The map distance is 8.4 cm. The actual distance is \(8.4 \times 25\,000 = 210\,000\text{ cm}\). Since there are 100 cm in a metre, this is \(210\,000 \div 100 = 2100\text{ m}\). Since there are 1000 m in a kilometre, this is \(2100 \div 1000 = 2.1\text{ km}\).

M1 for \(8.4 \times 25\,000\) or 210,000 or 2100. A1 for 2.1.

The formula for the volume of a cylinder is \(V = \pi r^2 h\). Substituting the given values: \(V = \pi \times 4^2 \times 9 = 144\pi\text{ cm}^3\). Using a calculator: \(144\pi \approx 452.389\text{ cm}^3\). Rounded to 1 decimal place, the volume is 452.4.

M1 for \(\pi \times 4^2 \times 9\) or \(144\pi\) or \(452.389\dots\) A1 for 452.4.

The sum of the probabilities of all outcomes must be 1. The probability of choosing a yellow counter is \(1 - 0.35 - 0.4 = 0.25\). Let \(N\) be the total number of counters in the bag. Then, \(0.25 \times N = 15\). Therefore, \(N = 15 \div 0.25 = 60\).

M1 for \(1 - 0.35 - 0.4 = 0.25\) or writing \(15 \div 0.25\). A1 for 60.

First find the total number of goals scored: \((0 \times 3) + (1 \times 7) + (2 \times 6) + (3 \times 4) = 0 + 7 + 12 + 12 = 31\text{ goals}\). The total number of matches is \(3 + 7 + 6 + 4 = 20\). The mean is \(31 \div 20 = 1.55\).

M1 for finding the total number of goals is 31, or showing a fully correct method for calculating the total goals. A1 for 1.55.

To find the error interval for a value rounded to the nearest \(5\text{ m}\), we find half of this degree of accuracy:
\(5 \div 2 = 2.5\text{ m}\).

Subtract \(2.5\text{ m}\) to find the lower bound:
\(70 - 2.5 = 67.5\text{ m}\)

Add \(2.5\text{ m}\) to find the upper bound:
\(70 + 2.5 = 72.5\text{ m}\)

The error interval is:
\(67.5 \le L < 72.5\)

M1 for identifying either boundary: \(67.5\) or \(72.5\) (or seen in an inequality)

A1 for \(67.5 \le L < 72.5\) (accept equivalent notation such as \([67.5, 72.5)\))

The total probability of all outcomes is \(1\).

First, find the probability of choosing either a blue or a yellow counter:
\(P(\text{blue or yellow}) = 1 - 0.45 = 0.55\)

The ratio of blue to yellow is \(3 : 8\).
Total number of parts in the ratio is \(3 + 8 = 11\).

The probability of choosing a blue counter is:
\(0.55 \times \frac{3}{11} = 0.05 \times 3 = 0.15\)

M1 for \(1 - 0.45\) (or \(0.55\)) or for dividing their remaining probability by \(11\)

A1 for \(0.15\) (or equivalent fraction, e.g., \(\frac{3}{20}\) or \(15\%\))

First, expand each bracket carefully, paying close attention to the signs:

\(4(3x - 2) = 12x - 8\)

\(-3(x - 5) = -3x + 15\)

Now, collect the like terms:

\(12x - 8 - 3x + 15 = 12x - 3x - 8 + 15\)

\(= 9x + 7\)

M1 for at least one correct expansion: \(12x - 8\) or \(-3x + 15\) (or \(-(3x - 15)\))

A1 for \(9x + 7\)

Let the initial valuation be \(V\).

After the first year (8% increase): \(V \times 1.08\)
After the second year (5% decrease): \(V \times 1.08 \times 0.95\)
After the third year (12% increase): \(V \times 1.08 \times 0.95 \times 1.12 = 344,736\)

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

Set up the equation:
\(1.14912 \times V = 344,736\)

Solve for \(V\):
\(V = \frac{344,736}{1.14912} = 300,000\)

The initial valuation was £300,000.

M1: For writing a correct equation or sequence of calculations representing the compounding changes, e.g., \(V \times 1.08 \times 0.95 \times 1.12 = 344,736\) or finding the combined multiplier \(1.14912\).
M1: For a correct division method, e.g., \(\frac{344,736}{1.14912}\) or sequential division.
A1.3: For the correct final answer of £300,000 (accept 300000).

First, calculate the area of the circle:
\(\text{Area of circle} = \pi \times r^2 = \pi \times 8^2 = 64\pi \approx 201.062\text{ cm}^2\)

Next, calculate the area of the regular hexagon. A regular hexagon of side length 8 cm consists of 6 equilateral triangles, each with side length 8 cm.
\(\text{Area of one equilateral triangle} = \frac{1}{2} \times a \times b \times \sin(C) = \frac{1}{2} \times 8 \times 8 \times \sin(60^\circ) = 32 \times \frac{\sqrt{3}}{2} = 16\sqrt{3} \approx 27.713\text{ cm}^2\)

\(\text{Total area of the hexagon} = 6 \times 16\sqrt{3} = 96\sqrt{3} \approx 166.277\text{ cm}^2\)

Calculate the remaining area:
\(\text{Remaining area} = 64\pi - 96\sqrt{3} \approx 201.062 - 166.277 = 34.785\text{ cm}^2\)

To 3 significant figures, this is \(34.8\text{ cm}^2\).

M1: For calculating the area of the circle: \(64\pi\) (or approx. 201.1 or 201.06)
M1: For finding the area of the regular hexagon: \(6 \times \frac{1}{2} \times 8^2 \times \sin(60^\circ)\) or \(96\sqrt{3}\) (or approx. 166.3)
A1.3: For subtracting the two areas to get \(34.8\) (accept 34.78 to 34.8).

Let \(H\) be the set of students who study History and \(G\) be the set of students who study Geography.

The total number of students in the union \(H \cup G\) is:
\(30 - 4 = 26\)

Using the principle of inclusion-exclusion:
\(n(H \cup G) = n(H) + n(G) - n(H \cap G)\)
\(26 = 18 + 15 - n(H \cap G)\)
\(26 = 33 - n(H \cap G)\)
\(n(H \cap G) = 33 - 26 = 7\)

So, 7 students study both History and Geography.

We want to find the conditional probability \(P(G | H)\):
\(P(G | H) = \frac{n(H \cap G)}{n(H)} = \frac{7}{18}\)

M1: For identifying that the number of students studying at least one subject is \(30 - 4 = 26\).
M1: For calculating the number of students studying both subjects: \(18 + 15 - 26 = 7\).
A1.3: For the correct probability of \(\frac{7}{18}\).

Identify the midpoints of each class interval:
- \(0 < t \le 10\): Midpoint = 5
- \(10 < t \le 20\): Midpoint = 15
- \(20 < t \le 30\): Midpoint = 25
- \(30 < t \le 40\): Midpoint = 35

Calculate the sum of products (Midpoint \(\times\) Frequency):
\(\sum f \times m = (5 \times 5) + (8 \times 15) + (x \times 25) + (4 \times 35)\)
\(\sum f \times m = 25 + 120 + 25x + 140 = 285 + 25x\)

Calculate the sum of frequencies:
\(\sum f = 5 + 8 + x + 4 = 17 + x\)

Set up the equation for the estimated mean:
\(\frac{285 + 25x}{17 + x} = 18\)

Solve for \(x\):
\(285 + 25x = 18(17 + x)\)
\(285 + 25x = 306 + 18x\)
\(7x = 306 - 285\)
\(7x = 21\)
\(x = 3\)

M1: For writing a correct expression for the total frequency \(17+x\) or the sum of products \(285+25x\) (at least 3 correct midpoints used).
M1: For forming the equation \(\frac{285 + 25x}{17 + x} = 18\) and showing algebraic steps to solve it.
A1.3: For obtaining \(x = 3\).

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

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

Combine the numerators over the common denominator:

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

Expand the terms in the numerator:

\(3(x-1) = 3x - 3\)
\(2(x+2) = 2x + 4\)

Subtract the expanded expressions:

\((3x - 3) - (2x + 4) = 3x - 3 - 2x - 4 = x - 7\)

Write the final fraction:

\(\frac{x-7}{(x+2)(x-1)}\) (or \(\frac{x-7}{x^2+x-2}\))

M1: For establishing a common denominator of \((x+2)(x-1)\).
M1: For correctly expanding the numerators: \(3x - 3\) and \(2x + 4\) (watch for the negative sign subtraction error).
A1.3: For the final correct fraction: \(\frac{x-7}{(x+2)(x-1)}\) or \(\frac{x-7}{x^2+x-2}\).

The formula for the area of a triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\).

Set up the equation:
\(\frac{1}{2}(2x - 3)(x + 4) = 20\)

Multiply both sides by 2:
\((2x - 3)(x + 4) = 40\)

Expand the brackets:
\(2x^2 + 8x - 3x - 12 = 40\)
\(2x^2 + 5x - 12 = 40\)

Rearrange into a quadratic form set to zero:
\(2x^2 + 5x - 52 = 0\)

Factorise the quadratic expression:
We need factors of \(2 \times (-52) = -104\) that add up to \(5\). These are \(13\) and \(-8\).
\(2x^2 - 8x + 13x - 52 = 0\)
\(2x(x - 4) + 13(x - 4) = 0\)
\((2x + 13)(x - 4) = 0\)

This gives two solutions:
\(x = -6.5\) or \(x = 4\)

Since \(x\) represents lengths, \(2x - 3\) and \(x + 4\) must be positive. Therefore, we discard \(x = -6.5\).

Thus, \(x = 4\).

M1: For setting up the area equation \(\frac{1}{2}(2x-3)(x+4) = 20\).
M1: For expanding and rearranging to the quadratic form \(2x^2 + 5x - 52 = 0\).
A1.3: For solving the quadratic and selecting the correct positive solution \(x = 4\).

Let \(P\) be the original price of the coat.

First reduction of 15% means the sale price is \(0.85P\).
An additional 10% off the sale price means the final price is \(0.85P \times 0.90\).

Calculate the overall multiplier:
\(0.85 \times 0.90 = 0.765\)

We know the final price is £53.55:
\(0.765P = 53.55\)

Solve for \(P\):
\(P = \frac{53.55}{0.765} = 70\)

The original price of the coat was £70.

M1: For setting up the overall percentage multiplier, e.g., \(0.85 \times 0.90\) or identifying \(76.5\%\) of the original price.
M1: For setting up the division \(\frac{53.55}{0.765}\) (or equivalent process of working backwards step-by-step).
A1.3: For the correct answer £70 (accept 70).

Step 1: Find the gradient of line \(L_1\):
\(m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2\)

Step 2: Find the gradient of line \(L_2\) (perpendicular to \(L_1\)):
\(m_2 = -\frac{1}{m_1} = -\frac{1}{2} = -0.5\)

Step 3: Find the coordinates of the midpoint of \(AB\):
\(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{2 + 6}{2}, \frac{5 + 13}{2}\right) = (4, 9)\)

Step 4: Use the point-slope form to find the equation of \(L_2\):
\(y - y_1 = m(x - x_1)\)
\(y - 9 = -0.5(x - 4)\)
\(y - 9 = -0.5x + 2\)
\(y = -0.5x + 11\)

M1: For finding the gradient of \(L_1\) to be \(2\) and correctly determining that the perpendicular gradient is \(-0.5\) (or \(-1/2\)).
M1: For finding the correct midpoint of \(AB\) as \((4, 9)\).
A1.3: For producing the final equation \(y = -0.5x + 11\) (or equivalent, e.g. \(y = -\frac{1}{2}x + 11\)).

To find the upper bound for the density, we use the formula: Density = Mass / Volume. To maximise this fraction, we need the upper bound of the mass and the lower bound of the volume. 1. Find the bounds for mass: Mass is 415 g to the nearest 5 g, so the upper bound is \(415 + 2.5 = 417.5\text{ g}\). 2. Find the bounds for volume: Volume is 80 cm\(^3\) to the nearest 10 cm\(^3\), so the lower bound is \(80 - 5 = 75\text{ cm}^3\). 3. Calculate the upper bound of density: \(\text{Density}_{\text{UB}} = 417.5 / 75 \approx 5.56667\text{ g/cm}^3\). Rounding to 3 significant figures gives \(5.57\text{ g/cm}^3\).

M1: For finding either the upper bound of mass (417.5) or the lower bound of volume (75). M1: For dividing their mass upper bound by their volume lower bound. A1: For 5.57.

1. Find the combined probability of picking a blue or a yellow counter: \(1 - 0.35 = 0.65\). 2. Use the ratio of blue to yellow counters (2 : 3) to find the fraction representing yellow: \(3 / (2 + 3) = 3/5 = 0.6\). 3. Calculate the probability of picking a yellow counter: \(0.65 \times 0.6 = 0.39\).

M1: For finding the remaining probability of 0.65. M1: For multiplying their remaining probability by 3/5. A1: For 0.39.

1. Find the area of the cross-section: Area of the rectangle = \(12 \times 6 = 72\text{ cm}^2\). Area of the semi-circle = \(0.5 \times \pi \times 3^2 = 4.5\pi \approx 14.137\text{ cm}^2\). Total cross-sectional area = \(72 + 14.137 = 86.137\text{ cm}^2\). 2. Calculate the volume of the prism: \(\text{Volume} = 86.137 \times 10 = 861.37\text{ cm}^3\). 3. Rounding to 3 significant figures: \(861\text{ cm}^3\).

M1: For calculating the area of the rectangle (72) OR the area of the semi-circle (4.5*pi or approx 14.1). M1: For finding the total cross-sectional area and multiplying by 10. A1: For 861 (accept 861 - 862).

1. Set up the equation for the area of the rectangle: \((x + 5)(x - 2) = 18\). 2. Expand and rearrange the equation to form a quadratic: \(x^2 + 3x - 10 = 18\) which becomes \(x^2 + 3x - 28 = 0\). 3. Solve the quadratic equation by factorisation: \((x + 7)(x - 4) = 0\). This gives \(x = -7\) or \(x = 4\). 4. Since lengths must be positive, \(x = -7\) is rejected. Thus, \(x = 4\).

M1: For setting up the equation \((x + 5)(x - 2) = 18\) and expanding to quadratic form. M1: For rearranging to \(x^2 + 3x - 28 = 0\) and attempting to factorise. A1: For 4 (must explicitly reject -7 or state 4 as the only final solution).

Let S be the stock value at the start of 2021. 1. Apply the 15% increase for 2021: value is \(1.15 \times S\). 2. Apply the 8% decrease for 2022: value is \(1.15 \times S \times 0.92 = 1.058 \times S\). 3. Set up the equation: \(1.058 \times S = 24380\). Solving for S gives \(S = 24380 / 1.058 = 23000\).

M1: For finding the combined multiplier of 1.058. M1: For setting up the division 24380 / 1.058. A1: For 23000.

1. Find the midpoints (x) for each interval: 5, 15, 25, and 35. 2. Multiply each midpoint by its frequency (f * x): \(6 \times 5 = 30\), \(14 \times 15 = 210\), \(15 \times 25 = 375\), \(5 \times 35 = 175\). 3. Sum these values: \(\sum f x = 30 + 210 + 375 + 175 = 790\). 4. Calculate the estimate of the mean: \(790 / 40 = 19.75\).

M1: For writing down the correct midpoints: 5, 15, 25, 35 (at least 3 correct shown or implied). M1: For summing (midpoint * frequency) to get 790. A1: For 19.75 (or 79/4).

1. Find the first differences: 7, 11, 15, 19. 2. Find the second differences: 4, 4, 4. Since the second difference is constant at 4, the coefficient of \(n^2\) is \(4 / 2 = 2\). 3. Subtract \(2n^2\) from each term of the original sequence: for n=1: \(4 - 2(1) = 2\); for n=2: \(11 - 8 = 3\); for n=3: \(22 - 18 = 4\); for n=4: \(37 - 32 = 5\); for n=5: \(56 - 50 = 6\). This leaves the linear sequence 2, 3, 4, 5, 6. 4. Find the n-th term of 2, 3, 4, 5, 6 which is \(n + 1\). 5. Combine the parts: \(2n^2 + n + 1\).

M1: For finding the constant second difference of 4 and dividing by 2 to get 2n^2. M1: For subtracting 2n^2 from terms of the sequence to obtain the linear sequence 2, 3, 4, 5, 6 and finding its n-th term (n + 1). A1: For 2n^2 + n + 1.

1. Set up the proportional equation: \(y = k / x^2\). 2. Find k using x = 3, y = 8: \(8 = k / 3^2 \implies 8 = k / 9 \implies k = 72\). 3. Use the formula \(y = 72 / x^2\) with x = 6: \(y = 72 / 6^2 = 72 / 36 = 2\).

M1: For setting up the proportional relationship \(y = k / x^2\). M1: For substituting x = 3 and y = 8 to find k = 72. A1: For 2.

First, express the numerator with a common power of 10: \(3.4 \times 10^5 + 0.42 \times 10^5 = 3.82 \times 10^5\). Next, divide by the denominator: \(\frac{3.82 \times 10^5}{2 \times 10^{-2}} = (3.82 \div 2) \times 10^{5 - (-2)} = 1.91 \times 10^7\).

B1 for correct option B chosen.

Since \(y \propto \frac{1}{x^2}\), we have \(y = \frac{k}{x^2}\). Substituting \(x = 4\) and \(y = 3\) gives \(3 = \frac{k}{4^2} = \frac{k}{16}\), so \(k = 48\). When \(x = 8\), \(y = \frac{48}{8^2} = \frac{48}{64} = 0.75\).

B1 for correct option B chosen.

The formula for the volume of a sphere is \(V = \frac{4}{3}\pi r^3\). Substituting \(r = 6\): \(V = \frac{4}{3} \times \pi \times 6^3 = \frac{4}{3} \times 216\pi = 288\pi\text{ cm}^3\).

B1 for correct option B chosen.

The number of students who study History but not Geography is \(18 - 6 = 12\). The total number of students is 30. Therefore, the probability is \(\frac{12}{30} = 0.4\).

B1 for correct option A chosen.

Factorise the numerator: \(2x^2 - 8 = 2(x^2 - 4) = 2(x-2)(x+2)\). Factorise the denominator: \(x^2 + 2x - 8 = (x+4)(x-2)\). Simplify by dividing both numerator and denominator by the common factor \(x-2\) to get \(\frac{2(x+2)}{x+4}\).

B1 for correct option C chosen.

First factorise the quadratic expression: \(x^2 - 5x - 14 = (x - 7)(x + 2) < 0\). The critical values are \(x = 7\) and \(x = -2\). For the product to be negative, \(x\) must lie between these critical values, so \(-2 < x < 7\).

B1 for correct option C chosen.

Rearranging \(L_1\) into \(y = mx + c\) form gives \(2y = 3x + 8 \implies y = \frac{3}{2}x + 4\). The gradient of \(L_1\) is \(\frac{3}{2}\). The gradient of the perpendicular line \(L_2\) is \(-\frac{2}{3}\). Using \(y - y_1 = m(x - x_1)\) with point \((6, 5)\): \(y - 5 = -\frac{2}{3}(x - 6) \implies y - 5 = -\frac{2}{3}x + 4 \implies y = -\frac{2}{3}x + 9\).

B1 for correct option B chosen.

Frequency is calculated as the class width multiplied by the frequency density. The class width is \(30 - 20 = 10\). Frequency = \(10 \times 2.4 = 24\).

B1 for correct option C chosen.

Since \(y\) is inversely proportional to the square of \(x\), we can write the relationship as:

\(y = \dfrac{k}{x^2}\)

Substitute the known values \(x = 4\) and \(y = 3\) to find the constant \(k\):

\(3 = \dfrac{k}{4^2}\)

\(3 = \dfrac{k}{16}\)

\(k = 48\)

So, the formula is:

\(y = \dfrac{48}{x^2}\)

Substitute \(x = 8\) into the formula to find \(y\):

\(y = \dfrac{48}{8^2} = \dfrac{48}{64} = 0.75\)

Alternatively, since \(x\) is multiplied by \(2\) (from \(4\) to \(8\)), and \(y\) is inversely proportional to \(x^2\), \(y\) must be divided by \(2^2 = 4\):

\(y = 3 \div 4 = 0.75\)

B1 for 0.75 (or equivalent fraction) or selecting Option B

To find the prime factors of 2340, we divide by prime numbers: \(2340 \div 2 = 1170\), \(1170 \div 2 = 585\), \(585 \div 3 = 195\), \(195 \div 3 = 65\), \(65 \div 5 = 13\), and 13 is prime. Thus, the prime factors are \(2 \times 2 \times 3 \times 3 \times 5 \times 13 = 2^2 \times 3^2 \times 5 \times 13\).

M1 for a complete and correct method to find prime factors (such as a factor tree with at least 3 correct branches, or a division ladder with at least 3 correct divisions). A1 for the correct index form \(2^2 \times 3^2 \times 5 \times 13\).

First, expand the brackets: \(7x - 4 > 3x + 24\). Next, subtract \(3x\) from both sides: \(4x - 4 > 24\). Add 4 to both sides: \(4x > 28\). Divide both sides by 4: \(x > 7\).

M1 for expanding the brackets correctly to get \(3x + 24\), or for correctly rearranging an incorrect expansion to group like terms. A1 for the correct inequality \(x > 7\).

The measurement of \(14.6\text{ cm}\) is correct to the nearest millimetre (\(0.1\text{ cm}\)). The maximum error is half of this degree of accuracy: \(0.1 \div 2 = 0.05\text{ cm}\). Therefore, the upper bound is \(14.6 + 0.05 = 14.65\text{ cm}\).

M1 for identifying the half-interval of \(0.05\text{ cm}\) or writing down \(14.65\) as a potential boundary. A1 for the correct upper bound of \(14.65\text{ cm}\) (or \(14.65\)).

We can write the relationship as \(y = k\sqrt{x}\). Substitute \(x = 36\) and \(y = 15\) into the equation: \(15 = k\sqrt{36}\), which simplifies to \(15 = 6k\). Solving for \(k\) gives \(k = 2.5\). Now, use \(k = 2.5\) and \(x = 81\) to find \(y\): \(y = 2.5\sqrt{81} = 2.5 \times 9 = 22.5\).

M1 for setting up the proportion equation \(y = k\sqrt{x}\) and solving for \(k\) to get \(2.5\) (or an equivalent ratio method). A1 for the correct final answer of \(22.5\).

The sum of the probabilities of all mutually exclusive outcomes is 1. Let the probability of choosing a yellow counter be \(x\). This means the probability of choosing a blue counter is \(3x\). We can set up the equation: \(0.36 + x + 3x = 1\). Simplifying gives \(0.36 + 4x = 1\), which means \(4x = 0.64\). Dividing by 4, we find \(x = 0.16\).

M1 for setting up a correct equation, e.g. \(0.36 + 4x = 1\), or for showing \(1 - 0.36 = 0.64\) and dividing by 4. A1 for the correct probability of \(0.16\).

To factorise fully, find the highest common factor of both terms. The highest common factor of \(12y^2\) and \(18y\) is \(6y\). Dividing both terms by \(6y\), we get \(6y(2y - 3)\).

M1 for factorising out any common factor, such as \(2y(6y - 9)\), \(3(4y^2 - 6y)\), or \(y(12y - 18)\). A1 for the fully factorised expression \(6y(2y - 3)\).

The formula for the area of a sector is \(\frac{\theta}{360} \times \pi r^2\). Substituting the given values: \(\text{Area} = \frac{45}{360} \times \pi \times 8^2 = \frac{1}{8} \times 64\pi = 8\pi\text{ cm}^2\). Using a calculator, \(8\pi \approx 25.1327...\text{ cm}^2\). Rounded to 3 significant figures, this is \(25.1\).

M1 for a correct substitution into the sector area formula, i.e., \(\frac{45}{360} \times \pi \times 8^2\). A1 for the correct answer rounded to 3 significant figures: \(25.1\).

An increase of \(15\%\) corresponds to a multiplier of \(1.15\). Let the original price be \(P\). We have the equation: \(1.15 \times P = 92\). Solving for \(P\): \(P = \frac{92}{1.15} = 80\). So the original price was \(\pounds 80\).

M1 for a division of 92 by 1.15, or setting up the equation \(1.15x = 92\) (or equivalent percentage method, e.g. finding that \(115\% = 92\)). A1 for the correct original price of \(80\) (or \(\pounds 80\)).

The total probability of choosing a red or blue counter is \(1 - 0.2 = 0.8\). The ratio of red to blue counters is \(3 : 5\), which gives a total of \(3 + 5 = 8\) parts. The probability of choosing a blue counter is therefore \(\frac{5}{8}\) of \(0.8\). Calculating this gives: \(\frac{5}{8} \times 0.8 = 0.5\).

M1 for \(1 - 0.2\) or \(0.8\) seen, or for dividing \(0.8\) by 8 (to get \(0.1\)). A1 for \(0.5\) (or equivalent fraction or percentage).

First, find the rate of bottling per minute: \(180 \div 4 = 45\) drinks per minute. Next, find the number of drinks bottled in 15 minutes: \(45 \times 15 = 675\) drinks. Alternatively, find the scaling factor for time: \(15 \div 4 = 3.75\), then multiply the number of drinks by this factor: \(180 \times 3.75 = 675\).

M1 for \(180 \div 4\) (or \(45\)) or \(15 \div 4\) (or \(3.75\)) or \(180 \times 15\) (or \(2700\)). A1 for \(675\).

Find the highest common factor of the terms \(6x^2 y\) and \(9xy^2\). The highest common factor of the coefficients 6 and 9 is 3. 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 place outside the bracket is \(3xy\). Divide each term by \(3xy\) to find the terms inside the bracket: \(6x^2 y \div 3xy = 2x\) and \(9xy^2 \div 3xy = 3y\). This gives the factorised expression: \(3xy(2x - 3y)\).

M1 for factorising out any common factor of at least degree 1 (e.g. \(3(2x^2 y - 3xy^2)\), \(xy(6x - 9y)\), or \(3x(2xy - 3y^2)\), or for a correct partial factorisation with a two-term expression in brackets). A1 for \(3xy(2x - 3y)\) or equivalent fully factorised form.

The side length \(s = 6.4\text{ cm}\) is given to 1 decimal place, so the degree of accuracy is \(0.1\text{ cm}\). The lower bound of the side length is \(6.4 - 0.05 = 6.35\text{ cm}\). The perimeter of a square is given by \(4s\). Therefore, the lower bound of the perimeter is \(4 \times 6.35 = 25.4\text{ cm}\).

M1 for identifying \(6.35\) as the lower bound of the side length, or for showing \(4 \times s_{\text{LB}}\) where \(6.3 \le s_{\text{LB}} < 6.4\). A1 for \(25.4\).

1. Calculate the rate of the first machine in litres per minute:
Total volume filled in 10 minutes = \(80 \times 250\text{ ml} = 20,000\text{ ml} = 20\text{ litres}\).
Rate = \(20\text{ litres} \div 10\text{ minutes} = 2\text{ litres per minute}\).

2. Calculate the rate of the second machine:
Rate = \(2\text{ litres per minute} \times 1.8 = 3.6\text{ litres per minute}\).

3. Calculate the time to fill 45 litres:
Time = \(45\text{ litres} \div 3.6\text{ litres per minute} = 12.5\text{ minutes}\).

M1: For finding the total volume filled by the first machine in litres, e.g., \(20\text{ litres}\) or finding its rate, e.g., \(2\text{ litres/min}\)
M1: For multiplying their rate by 1.8 to get \(3.6\text{ litres/min}\) (or equivalent method with total bottles)
A1: For 12.5 (accept 12 minutes 30 seconds)

The area of the triangular cross-section is:
\[\text{Area} = \frac{1}{2} \times x \times (x + 3)\]

The volume of the prism is the area of the cross-section multiplied by its length:
\[\text{Volume} = \frac{1}{2} x (x + 3) \times 10 = 5x(x + 3)\]

We are given that the volume is \(540\text{ cm}^3\):
\[5x(x + 3) = 540\]
\[x(x + 3) = 108\]
\[x^2 + 3x - 108 = 0\]

Factorising the quadratic equation:
\[(x + 12)(x - 9) = 0\]

Since \(x\) must be a positive length, we choose \(x = 9\).

M1: For setting up an equation for the volume of the prism, e.g., \(\frac{1}{2} x (x + 3) \times 10 = 540\)
M1: For simplifying to a quadratic equation, e.g., \(x^2 + 3x - 108 = 0\) or solving by inspection from \(x(x+3) = 108\)
A1: For 9

To eliminate \(y\), multiply the first equation by 2 and the second equation by 3:
\[8x - 6y = 36\]
\[15x + 6y = 33\]

Add the two equations:
\[(8x + 15x) = 36 + 33\]
\[23x = 69\]
\[x = 3\]

Substitute \(x = 3\) into the second equation:
\[5(3) + 2y = 11\]
\[15 + 2y = 11\]
\[2y = -4\]
\[y = -2\]

So the solution is \(x = 3\), \(y = -2\).

M1: For a correct method to eliminate one variable, e.g., multiplying equations to get coefficients of \(y\) to be \(-6\) and \(6\), or coefficients of \(x\) to be \(20\) and \(-20\)
M1: For finding the correct value of one variable (either \(x = 3\) or \(y = -2\))
A1: For both \(x = 3\) and \(y = -2\)

1. Find the total score for the boys:
\[12 \times 65 = 780\]

2. Find the total score for the girls:
\[18 \times 72 = 1296\]

3. Find the total score for all students:
\[780 + 1296 = 2076\]

4. Calculate the overall mean:
\[2076 \div 30 = 69.2\]

M1: For finding the total score of the boys (780) or the total score of the girls (1296)
M1: For finding the combined total score (2076) and dividing by 30
A1: For 69.2

1. Find the probability of choosing either a blue or a yellow counter:
\[P(\text{Blue or Yellow}) = 1 - 0.3 = 0.7\]

2. The ratio of blue to yellow is \(3 : 4\), so the yellow counters represent \(\frac{4}{3 + 4} = \frac{4}{7}\) of the non-red counters.

3. Calculate the probability of choosing a yellow counter:
\[P(\text{Yellow}) = \frac{4}{7} \times 0.7 = 0.4\]

M1: For finding the probability of not red, i.e., \(1 - 0.3 = 0.7\)
M1: For a correct method to divide 0.7 in the ratio \(3 : 4\), e.g., \(0.7 \div (3 + 4) \times 4\)
A1: For 0.4 (or equivalent fraction, e.g., \(\frac{2}{5}\))

1. Multiply both sides by \((x + 2)\):
\[y(x + 2) = 5x - 3\]
\[xy + 2y = 5x - 3\]

2. Rearrange the terms to get all terms with \(x\) on one side:
\[2y + 3 = 5x - xy\]

3. Factorise \(x\) on the right-hand side:
\[2y + 3 = x(5 - y)\]

4. Divide by \((5 - y)\):
\[x = \frac{2y + 3}{5 - y}\]

M1: For multiplying by \((x + 2)\) and expanding the bracket correctly, e.g., \(xy + 2y = 5x - 3\)
M1: For isolating terms in \(x\) on one side and factorising, e.g., \(x(5 - y) = 2y + 3\) (or equivalent)
A1: For \(x = \frac{2y + 3}{5 - y}\) (or equivalent, e.g., \(x = \frac{-2y - 3}{y - 5}\))

Let the original price of the coat be \(P\).

1. After a 20% reduction, the price is:
\[0.80 \times P\]

2. After a further 15% reduction, the price is:
\[0.80 \times P \times 0.85 = 0.68 \times P\]

3. We are given the final price is £102:
\[0.68 \times P = 102\]
\[P = \frac{102}{0.68} = 150\]

So, the original price of the coat was £150.

M1: For a correct multiplier for 20% reduction (0.8) or 15% reduction (0.85)
M1: For setting up the equation \(0.8 \times 0.85 \times P = 102\) or \(0.68 \times P = 102\) (or working backwards: \(102 \div 0.85 = 120\))
A1: For 150 (or £150)

1. Find the number of sides, \(n\), using the sum of interior angles formula:
\[(n - 2) \times 180^\circ = 1800^\circ\]
\[n - 2 = \frac{1800}{180} = 10\]
\[n = 12\]

So the polygon has 12 sides.

2. Work out the size of one exterior angle:
\[\text{Exterior angle} = \frac{360^\circ}{n} = \frac{360^\circ}{12} = 30^\circ\]

Alternatively:
One interior angle = \(1800^\circ \div 12 = 150^\circ\).
Exterior angle = \(180^\circ - 150^\circ = 30^\circ\).

M1: For setting up a correct equation to find the number of sides, e.g., \((n-2) \times 180 = 1800\)
M1: For finding the number of sides \(n = 12\) or finding the size of one interior angle as \(150^\circ\)
A1: For 30 (accept \(30^\circ\))

First, identify the dimensions of the smallest face. The dimensions are 30 cm, 80 cm, and 1.2 m. Convert these dimensions to metres: 30 cm = 0.3 m and 80 cm = 0.8 m. The area of the smallest face is \(0.3 \times 0.8 = 0.24\) square metres. Use the formula for pressure: \(\text{Pressure} = \text{Force} / \text{Area}\). This gives \(\text{Pressure} = 144 / 0.24 = 600\text{ N/m}^2\).

M1: For converting dimensions to metres: 0.3 m and 0.8 m (or finding the area in \(cm^2\) as \(2400\text{ cm}^2\)). M1: For calculating the area of the smallest face as \(0.24\text{ m}^2\) (or converting \(2400\text{ cm}^2\) to \(m^2\) by dividing by 10000). A1: For 600.

Let the original price be P. After a 15% reduction, the price is \(P \times 0.85\). After a further 10% reduction, the price is \(P \times 0.85 \times 0.90 = 0.765 P\). Set up the equation: \(0.765 P = 61.20\). Solve for P: \(P = 61.20 / 0.765 = 80\).

M1: For finding a combined multiplier of \(0.85 \times 0.9 = 0.765\) (or equivalent). M1: For setting up the equation \(0.765 \times P = 61.20\) or calculating \(61.20 / 0.9 = 68\). A1: For 80.

The volume of a cylinder is \(\pi r^2 h\). Volume of cylinder = \(\pi \times 4^2 \times 12 = 192\pi \approx 603.18578\text{ cm}^3\). The volume of a sphere is \((4/3)\pi R^3\). Since they are equal: \((4/3)\pi R^3 = 192\pi\). Simplify to find \(R^3\): \(R^3 = 192 \times (3/4) = 144\). Find the cube root of 144: \(R = \sqrt[3]{144} \approx 5.24148\text{ cm}\). To 3 significant figures, this is 5.24.

M1: For finding the volume of the cylinder as \(192\pi\) or approximately 603. M1: For equating the volume formulas and finding \(R^3 = 144\). A1: For 5.24 (accept answers in the range 5.24 to 5.242).

Set up an equation for the area: \((2x + 3)(x + 1) = 45\). Expand the brackets: \(2x^2 + 2x + 3x + 3 = 45\), which simplifies to \(2x^2 + 5x + 3 = 45\). Rearrange into standard quadratic form: \(2x^2 + 5x - 42 = 0\). Factorise the quadratic equation: \((2x - 7)(x + 6) = 0\). This gives \(x = 3.5\) or \(x = -6\). Since length must be positive, we reject \(x = -6\), so \(x = 3.5\).

M1: For setting up the equation \((2x + 3)(x + 1) = 45\) and expanding to \(2x^2 + 5x + 3 = 45\). M1: For rearranging to \(2x^2 + 5x - 42 = 0\) and attempting to solve by factorisation or formula. A1: For 3.5 (accept 7/2).

The total number of counters in the bag is \(n + 4\). The probability of choosing a red counter first is \(4 / (n + 4)\). The probability of choosing a red counter second, without replacement, is \(3 / (n + 3)\). The probability of both being red is \((4 / (n + 4)) \times (3 / (n + 3)) = 12 / ((n + 4)(n + 3))\). Set this equal to 2/15: \(12 / ((n + 4)(n + 3)) = 2 / 15\). Simplify: \(2(n + 4)(n + 3) = 180 \implies (n + 4)(n + 3) = 90\). Expand and solve: \(n^2 + 7n + 12 = 90 \implies n^2 + 7n - 78 = 0\). Factorise: \((n - 6)(n + 13) = 0\). Since n must be positive, \(n = 6\).

M1: For writing an expression for the probability of selecting two red counters, e.g., \(\frac{4}{n+4} \times \frac{3}{n+3}\). M1: For setting up the equation and simplifying to a quadratic equation, e.g., \(n^2 + 7n - 78 = 0\). A1: For 6.

To find the population density, divide the total population by the total area: \(\text{Density} = (6.4 \times 10^7) / (2.5 \times 10^5)\). Divide the numbers: \(6.4 / 2.5 = 2.56\). Divide the powers of 10: \(10^7 / 10^5 = 10^2\). This gives \(2.56 \times 10^2\) people per \(\text{km}^2\).

M1: For identifying the correct division needed: \((6.4 \times 10^7) / (2.5 \times 10^5)\). M1: For calculating 256 or showing \(2.56 \times 10^n\). A1: For \(2.56 \times 10^2\) (or \(2.56 \times 10^2\)).

Multiply both sides by \((2 - x)\) to clear the denominator: \(y(2 - x) = 3x + 5\). Expand the bracket: \(2y - xy = 3x + 5\). Move all terms with x to one side and other terms to the opposite side: \(2y - 5 = 3x + xy\). Factorise x from the terms on the right side: \(2y - 5 = x(3 + y)\). Divide both sides by \((3 + y)\) to solve for x: \(x = \frac{2y - 5}{3 + y}\).

M1: For multiplying by \((2 - x)\) to get \(y(2 - x) = 3x + 5\). M1: For expanding and collecting the terms in x on one side: e.g., \(xy + 3x = 2y - 5\). A1: For \(x = \frac{2y - 5}{y + 3}\) (or equivalent).