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An increase of 12% represents a multiplier of 1.12. Original value = \( 313.60 \div 1.12 = 280 \). Thus, the original value was £280.

M1 for \( 313.60 \div 1.12 \) or an equivalent method (e.g., setting up \( 112\% = 313.60 \)). A1 for 280 (or £280).

First, find the reciprocal to eliminate the negative power: \( \left(\frac{64}{27}\right)^{\frac{2}{3}} \). Next, apply the cube root: \( \left(\sqrt[3]{\frac{64}{27}}\right)^2 = \left(\frac{4}{3}\right)^2 \). Finally, square the numerator and denominator: \( \frac{16}{9} \).

M1 for applying either the reciprocal or the cube root correctly, e.g. writing \( \left(\frac{64}{27}\right)^{\frac{2}{3}} \) or \( \left(\frac{3}{4}\right)^{-2} \). A1 for \( \frac{16}{9} \) (or \( 1\frac{7}{9} \)).

The lower bound for the length is \( 34.5\text{ m} \). The lower bound for the width is \( 19.5\text{ m} \). The lower bound for the area is \( 34.5 \times 19.5 = 672.75\text{ m}^2 \).

M1 for identifying at least one lower bound correctly (length \( = 34.5 \) or width \( = 19.5 \)) or for multiplying two bounds. A1 for 672.75 (accept \( 672.75\text{ m}^2 \)).

First, evaluate \( g(4) \): \( g(4) = 4^2 + 2 = 16 + 2 = 18 \). Now, substitute this result into \( f(x) \): \( f(18) = 3(18) - 5 = 54 - 5 = 49 \).

M1 for evaluating \( g(4) = 18 \) or expressing \( fg(x) = 3(x^2 + 2) - 5 \). A1 for 49.

Expand the bracket: \( 4x - 12 > 7x + 3 \). Subtract \( 4x \) from both sides: \( -12 > 3x + 3 \). Subtract 3 from both sides: \( -15 > 3x \). Divide by 3: \( -5 > x \), which is equivalent to \( x < -5 \).

M1 for expanding brackets and attempting to collect like terms on opposite sides, e.g. \( 4x - 12 > 7x + 3 \) leading to \( -3x > 15 \) or \( 3x < -15 \). A1 for \( x < -5 \) (or equivalent, e.g. \( -5 > x \)).

The difference in the number of parts between blue and red is \( 7 - 3 = 4 \) parts. This difference of 4 parts corresponds to 48 counters. Therefore, 1 part represents \( 48 \div 4 = 12 \) counters. The total number of parts is \( 3 + 7 = 10 \) parts. The total number of counters is \( 10 \times 12 = 120 \).

M1 for finding the value of one part: \( 48 \div (7 - 3) = 12 \), or setting up an equation like \( 7x - 3x = 48 \). A1 for 120.

Since the line is parallel to \( y = 4x - 3 \), its gradient \( m \) must be \( 4 \). The equation is of the form \( y = 4x + c \). Substitute the coordinates of the point \( (2, 10) \) to find \( c \): \( 10 = 4(2) + c \) which gives \( 10 = 8 + c \), so \( c = 2 \). Therefore, the equation of the line is \( y = 4x + 2 \).

M1 for identifying the gradient as 4 and substituting \( (2, 10) \) into \( y = 4x + c \) or \( y - 10 = 4(x - 2) \). A1 for \( y = 4x + 2 \) (or an equivalent form).

First, find the probability that the spinner lands on blue or red: \( P(\text{blue or red}) = 0.35 + 0.4 = 0.75 \). Then, subtract this from 1 to find the probability of not landing on blue or red: \( P(\text{not blue or red}) = 1 - 0.75 = 0.25 \).

M1 for \( 0.35 + 0.4 = 0.75 \) or for calculating \( 1 - 0.35 - 0.4 \). A1 for 0.25 (or equivalent fraction, e.g. \( \frac{1}{4} \)).

Let the annual multiplier be \(m\). The time period from the start of 2020 to the start of 2022 is 2 years. We can set up the equation: \(2400 \times m^2 = 2521.50\). Dividing both sides by 2400 gives: \(m^2 = \frac{2521.50}{2400} = 1.050625\). Taking the square root of both sides gives: \(m = \sqrt{1.050625} = 1.025\). A multiplier of 1.025 represents a percentage increase of \(2.5\%\). Thus, \(x = 2.5\).

M1 for setting up the equation \(2400 \times m^2 = 2521.50\) or finding \(\sqrt{\frac{2521.50}{2400}}\). A1 for \(2.5\).

Let the annual multiplier be \(m\). The time period from the start of 2020 to the start of 2022 is 2 years. We can set up the equation: \(2400 \times m^2 = 2521.50\). Dividing both sides by 2400 gives: \(m^2 = \frac{2521.50}{2400} = 1.050625\). Taking the square root of both sides gives: \(m = \sqrt{1.050625} = 1.025\). A multiplier of 1.025 represents a percentage increase of \(2.5\%\). Thus, \(x = 2.5\).

M1 for setting up the equation \(2400 \times m^2 = 2521.50\) or finding \(\sqrt{\frac{2521.50}{2400}}\). A1 for \(2.5\).

Let the annual multiplier be \(m\). The time period from the start of 2020 to the start of 2022 is 2 years. We can set up the equation: \(2400 \times m^2 = 2521.50\). Dividing both sides by 2400 gives: \(m^2 = \frac{2521.50}{2400} = 1.050625\). Taking the square root of both sides gives: \(m = \sqrt{1.050625} = 1.025\). A multiplier of 1.025 represents a percentage increase of \(2.5\%\). Thus, \(x = 2.5\).

M1 for setting up the equation \(2400 \times m^2 = 2521.50\) or finding \(\sqrt{\frac{2521.50}{2400}}\). A1 for \(2.5\).

Let the annual multiplier be \(m\). The time period from the start of 2020 to the start of 2022 is 2 years. We can set up the equation: \(2400 \times m^2 = 2521.50\). Dividing both sides by 2400 gives: \(m^2 = \frac{2521.50}{2400} = 1.050625\). Taking the square root of both sides gives: \(m = \sqrt{1.050625} = 1.025\). A multiplier of 1.025 represents a percentage increase of \(2.5\%\). Thus, \(x = 2.5\).

M1 for setting up the equation \(2400 \times m^2 = 2521.50\) or finding \(\sqrt{\frac{2521.50}{2400}}\). A1 for \(2.5\).

Let the annual multiplier be \(m\). The time period from the start of 2020 to the start of 2022 is 2 years. We can set up the equation: \(2400 \times m^2 = 2521.50\). Dividing both sides by 2400 gives: \(m^2 = \frac{2521.50}{2400} = 1.050625\). Taking the square root of both sides gives: \(m = \sqrt{1.050625} = 1.025\). A multiplier of 1.025 represents a percentage increase of \(2.5\%\). Thus, \(x = 2.5\).

M1 for setting up the equation \(2400 \times m^2 = 2521.50\) or finding \(\sqrt{\frac{2521.50}{2400}}\). A1 for \(2.5\).

Solve the first inequality:
\(3x - 5 > 7 \implies 3x > 12 \implies x > 4\).

Solve the second inequality:
\(x^2 - 12x + 32 \le 0\)
Factorising the quadratic expression:
\((x - 4)(x - 8) \le 0\)
The critical values are \(x = 4\) and \(x = 8\).
Since we require the expression to be less than or equal to 0, the variable must lie between these values:
\(4 \le x \le 8\).

Combining both inequalities:
We need values of \(x\) such that \(x > 4\) AND \(4 \le x \le 8\).
This yields the interval: \(4 < x \le 8\).

Identifying the integer values in this range:
Since \(x\) must be strictly greater than 4, the possible integers are \(5, 6, 7, 8\).

M1 for solving \(3x - 5 > 7\) to get \(x > 4\)
M1 for factorising \(x^2 - 12x + 32\) or finding critical values 4 and 8
M1 for obtaining the combined range \(4 < x \le 8\)
A1.5 for identifying the correct integers: 5, 6, 7, 8 (all four required, no extra values)

Let us determine the interior angle \(\angle PQR\) first.
The bearing of \(Q\) from \(P\) is \(040^\circ\).
This means the bearing of \(P\) from \(Q\) (backwards) is \(040^\circ + 180^\circ = 220^\circ\).
The bearing of \(R\) from \(Q\) is given as \(110^\circ\).
Therefore, the interior angle \(\angle PQR = 220^\circ - 110^\circ = 110^\circ\).

Now we have triangle \(PQR\) with:
\(PQ = 12\text{ km}\)
\(QR = 15\text{ km}\)
\(\angle PQR = 110^\circ\)

To find the distance \(PR\), we apply the Cosine Rule:
\(PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(\angle PQR)\)
\(PR^2 = 12^2 + 15^2 - 2 \times 12 \times 15 \times \cos(110^\circ)\)
\(PR^2 = 144 + 225 - 360 \times \cos(110^\circ)\)
\(PR^2 = 369 - 360 \times (-0.34202...)\)
\(PR^2 = 369 + 123.127...\)
\(PR^2 = 492.127...\)
\(PR = \sqrt{492.127...} \approx 22.1839\text{ km}\).

To 3 significant figures, the distance is \(22.2\text{ km}\).

M1 for finding the correct interior angle \(\angle PQR = 110^\circ\)
M1 for substituting correct values into the Cosine Rule: \(12^2 + 15^2 - 2 \times 12 \times 15 \times \cos(110^\circ)\)
M1 for evaluating to find \(PR^2 \approx 492.1\)
A1.5 for \(22.2\) (accept answers in the range 22.18 to 22.2)

The relationship is given by:
\(y = \frac{k}{\sqrt{x}}\)
where \(k\) is a constant.

Substitute the given values \(x = 16\) and \(y = 7.5\) to calculate \(k\):
\(7.5 = \frac{k}{\sqrt{16}}\)
\(7.5 = \frac{k}{4}\)
\(k = 7.5 \times 4 = 30\)

Now, rewrite the formula with the constant of proportion:
\(y = \frac{30}{\sqrt{x}}\)

Substitute \(x = 0.64\) into the formula:
\(y = \frac{30}{\sqrt{0.64}}\)
\(y = \frac{30}{0.8}\)
\(y = 37.5\)

M1 for writing the relationship as \(y = \frac{k}{\sqrt{x}}\)
M1 for substituting \(x = 16\) and \(y = 7.5\) and finding \(k = 30\)
M1 for substituting \(x = 0.64\) into their equation
A1.5 for \(37.5\)

Let \(n\) be the total number of counters originally in the bag.
Since the probability of choosing a red counter is \(0.3\), the original number of red counters must be \(0.3n\).

After a red counter is chosen first and not replaced:
- The number of red counters remaining is \(0.3n - 1\)
- The total number of counters remaining is \(n - 1\)

The probability of choosing two red counters is:
\(P(\text{Red, Red}) = P(\text{Red}_1) \times P(\text{Red}_2 | \text{Red}_1)\)
\(0.3 \times \frac{0.3n - 1}{n - 1} = \frac{4}{45}\)

Now, solve this equation for \(n\):
\(\frac{3}{10} \times \frac{0.3n - 1}{n - 1} = \frac{4}{45}\)
Multiply both sides by 10:
\(\frac{3(0.3n - 1)}{n - 1} = \frac{40}{45} = \frac{8}{9}\)
Multiply both sides by \(9(n - 1)\):
\(27(0.3n - 1) = 8(n - 1)\)
\(8.1n - 27 = 8n - 8\)
\(0.1n = 19\)
\(n = 190\)

M1 for expressing the number of red counters as \(0.3n\) (or using a fraction equivalent)
M1 for setting up the probability equation: \(0.3 \times \frac{0.3n - 1}{n - 1} = \frac{4}{45}\)
M1 for algebraic simplification to a linear equation, e.g., \(8.1n - 27 = 8n - 8\)
A1.5 for \(190\)

Find the value of the house at the end of the first year after the \(4\%\) increase:
\(\text{Value after 1 year} = 240,000 \times 1.04 = 249,600\)

In the second year, the value decreases by \(x\%\).
Let the multiplier for this decrease be \(m = 1 - \frac{x}{100}\).
\(249,600 \times m = 233,376\)

Solve for \(m\):
\(m = \frac{233,376}{249,600} = 0.935\)

Since \(1 - \frac{x}{100} = 0.935\):
\(\frac{x}{100} = 1 - 0.935 = 0.065\)
\(x = 6.5\)

Thus, the percentage decrease in the second year is \(6.5\%\).

M1 for calculating the value after 1 year: \(240,000 \times 1.04 = 249,600\)
M1 for setting up the equation for the second year's decay: \(249,600 \times (1 - \frac{x}{100}) = 233,376\)
M1 for finding the multiplier \(0.935\) or the absolute change \(16,224\) and dividing by \(249,600\)
A1.5 for \(6.5\)

Let the \(n\)-th term of the quadratic sequence be given by:
\(u_n = an^2 + bn + c\)

Given terms:
\(u_3 = 9a + 3b + c = 20\)
\(u_4 = 16a + 4b + c = 35\)
\(u_5 = 25a + 5b + c = 54\)

First differences:
\(u_4 - u_3 = 7a + b = 15\)
\(u_5 - u_4 = 9a + b = 19\)

Second difference:
\((9a + b) - (7a + b) = 2a = 4 \implies a = 2\).

Substitute \(a = 2\) into \(7a + b = 15\):
\(7(2) + b = 15 \implies b = 1\).

Substitute \(a = 2\) and \(b = 1\) into \(9a + 3b + c = 20\):
\(18 + 3 + c = 20 \implies c = -1\).

The expression for the \(n\)-th term is:
\(u_n = 2n^2 + n - 1\)

To find which term has a value of 299, set \(u_n = 299\):
\(2n^2 + n - 1 = 299 \implies 2n^2 + n - 300 = 0\)

Solve the quadratic equation using the quadratic formula:
\(n = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-300)}}{4} = \frac{-1 \pm \sqrt{2401}}{4} = \frac{-1 \pm 49}{4}\)

Since \(n\) must be a positive integer, we select the positive root:
\(n = \frac{48}{4} = 12\).

M1 for a method to find the second difference to get \(a = 2\)
M1 for substituting \(a\) back to find \(b = 1\) and \(c = -1\) (resulting in \(2n^2 + n - 1\))
M1 for setting up the quadratic equation \(2n^2 + n - 300 = 0\)
A1.5 for \(12\)

Let the linear scale factor from cone A to cone B be \(k\).

The ratio of their total surface areas is:
\(\frac{\text{Area}_B}{\text{Area}_A} = \frac{162\pi}{72\pi} = \frac{162}{72} = \frac{9}{4}\)

Since the area scale factor is \(k^2\), we find the linear scale factor \(k\) by taking the square root:
\(k^2 = \frac{9}{4} \implies k = \frac{3}{2} = 1.5\)

The volume scale factor is \(k^3\):
\(\frac{\text{Volume}_B}{\text{Volume}_A} = k^3 = \left(\frac{3}{2}\right)^3 = \frac{27}{8}\)

We are given that the volume of cone B is \(486\pi\text{ cm}^3\).
Therefore, the volume of cone A is:
\(\text{Volume}_A = \frac{\text{Volume}_B}{k^3} = 486\pi \times \frac{8}{27} = 18 \times 8 \times \pi = 144\pi\text{ cm}^3\).

M1 for finding the ratio of areas: \(\frac{162}{72}\) or \(\frac{9}{4}\)
M1 for finding the linear scale factor \(k = \sqrt{\frac{9}{4}} = 1.5\)
M1 for cubing the scale factor to find the volume ratio \(k^3 = \frac{27}{8}\) (or 3.375)
A1.5 for \(144\pi\) (accept 144pi, reject decimal approximations like 452.39)

Substitute the expression for \(y\) from the first equation into the second equation:
\(x^2 + (2x - 1)^2 = 13\)

Expand the quadratic expression:
\(x^2 + 4x^2 - 4x + 1 = 13\)
\(5x^2 - 4x + 1 = 13\)

Rearrange to form a quadratic equation equal to 0:
\(5x^2 - 4x - 12 = 0\)

Factorise the quadratic:
\((5x + 6)(x - 2) = 0\)

This gives the possible values for \(x\):
\(x = 2\) or \(x = -1.2\)

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

For \(x = 2\):
\(y = 2(2) - 1 = 3\)
This gives the coordinate pair \((2, 3)\).

For \(x = -1.2\):
\(y = 2(-1.2) - 1 = -2.4 - 1 = -3.4\)
This gives the coordinate pair \((-1.2, -3.4)\).

So the two solutions are \((2, 3)\) and \((-1.2, -3.4)\).

M1 for substituting \(y = 2x - 1\) into the circle equation to get \(x^2 + (2x - 1)^2 = 13\)
M1 for expanding and simplifying to a three-term quadratic: \(5x^2 - 4x - 12 = 0\)
M1 for solving the quadratic to find \(x = 2\) and \(x = -1.2\)
A1.5 for both correct coordinate pairs: \((2, 3)\) and \((-1.2, -3.4)\)

Let \(x\) be the annual multiplier for the compound interest.

After 2 years, the value of the investment is:
\[5000 \times x^2 = 5325.12\]

Solving for \(x^2\):
\[x^2 = \frac{5325.12}{5000} = 1.065024\]

Finding the annual multiplier \(x\):
\[x = \sqrt{1.065024} = 1.032\]
(This corresponds to an interest rate of \(3.2\%\) per annum).

To find the value of the investment after a further 3 years (which represents 5 years in total from the initial investment):
\[\text{Value} = 5325.12 \times 1.032^3\]
\[\text{Value} = 5325.12 \times 1.099114176 = 5852.884576...\]

Alternatively, using the 5-year period directly:
\[\text{Value} = 5000 \times 1.032^5 = 5852.884576...\]

Rounding to the nearest penny gives \(\pounds 5852.88\).

**M1** for setting up the equation for 2 years: \(5000 \times x^2 = 5325.12\) or equivalent.
**M1** for finding the annual multiplier \(x = 1.032\) (or identifying the interest rate as \(3.2\%\)).
**M1** for a method to calculate the final amount after a further 3 years, e.g., \(5325.12 \times 1.032^3\) or \(5000 \times 1.032^5\).
**A1** for \(5852.88\) (accept \(\pounds 5852.88\)).

First, we use the formula for the area of a triangle:
\[\text{Area} = \frac{1}{2} a c \sin(B)\]

Substitute the given values into the formula:
\[35 = \frac{1}{2} \times 11.2 \times 7.5 \times \sin(B)\]
\[35 = 42 \sin(B)\]
\[\sin(B) = \frac{35}{42} = \frac{5}{6}\]

Since angle \(ABC\) is obtuse (between \(90^\circ\) and \(180^\circ\)):
\[B = 180^\circ - \arcsin\left(\frac{5}{6}\right) \approx 180^\circ - 56.443^\circ = 123.557^\circ\]

Now, we use the Cosine Rule to find the length of \(AC\):
\[AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(B)\]
\[AC^2 = 7.5^2 + 11.2^2 - 2 \times 7.5 \times 11.2 \times \cos(123.557^\circ)\]
\[AC^2 = 56.25 + 125.44 - 168 \times (-0.55277)\]
\[AC^2 = 181.69 + 92.865 = 274.555\]
\[AC = \sqrt{274.555} \approx 16.5697\text{ cm}\]

Correct to 3 significant figures, the length of \(AC\) is \(16.6\text{ cm}\).

**M1** for substituting known values into the area formula: \(35 = 0.5 \times 7.5 \times 11.2 \times \sin(B)\) or \(\sin(B) = \frac{5}{6}\).
**M1** for finding the obtuse angle \(B \approx 123.6^\circ\) (accept angles in range \(123.5^\circ\) to \(124.0^\circ\)).
**M1** for substituting their angle \(B\) and given side lengths into the Cosine Rule: \(AC^2 = 7.5^2 + 11.2^2 - 2 \times 7.5 \times 11.2 \times \cos(B)\).
**M1** for evaluating \(AC^2 \approx 274.6\) (or \(AC \approx \sqrt{274.6}\)).
**A1** for \(16.6\) (accept answers in the range \(16.5\) to \(16.7\)).

Since \(y\) is inversely proportional to the square of \(x\), we can write the relationship as:
\[y = \frac{k}{x^2}\]
where \(k\) is a constant.

Using the first set of conditions, when \(x = d\), \(y = 12\):
\[12 = \frac{k}{d^2} \implies k = 12d^2\]

Using the second set of conditions, when \(x = d + 3\), \(y = 3\):
\[3 = \frac{k}{(d + 3)^2} \implies k = 3(d + 3)^2\]

Equating both expressions for \(k\):
\[12d^2 = 3(d + 3)^2\]

Divide both sides by 3:
\[4d^2 = (d + 3)^2\]

Since \(d\) is a positive constant, both sides represent positive values. We can take the square root of both sides:
\[2d = d + 3\]

Solving for \(d\):
\[d = 3\]

(Note: The alternative negative root gives \(2d = -(d + 3) \implies 3d = -3 \implies d = -1\), which is rejected since \(d\) must be positive).

**M1** for writing a correct general relationship, e.g., \(y = \frac{k}{x^2}\) or \(y x^2 = k\).
**M1** for setting up an equation in terms of \(d\), e.g., \(12d^2 = 3(d + 3)^2\) or \(\frac{12}{3} = \frac{(d+3)^2}{d^2}\).
**M1** for simplifying the equation to \(4d^2 = (d+3)^2\) and taking the square root to get \(2d = d+3\) (or expanding and solving the quadratic equation \(3d^2 - 6d - 9 = 0\)).
**A1** for \(3\) (with positive constant condition satisfied).

We can write \(27^{-\frac{2}{3}}\) as \(\frac{1}{27^{\frac{2}{3}}}\). First, evaluate the cube root: \(27^{\frac{1}{3}} = \sqrt[3]{27} = 3\). Then, square the result: \(3^2 = 9\). This gives \(\frac{1}{9}\).

M1 for expressing with a positive index as \(\frac{1}{27^{2/3}}\) or finding the cube root \(27^{1/3} = 3\). A1 for \(1/9\).

First expand the right side: \(5x - 7 < 3x + 15\). Subtract \(3x\) from both sides to get \(2x - 7 < 15\). Add 7 to both sides to get \(2x < 22\). Divide by 2 to get \(x < 11\).

M1 for correct expansion of brackets to \(3x + 15\) or for isolating the x terms on one side. A1 for \(x < 11\).

Write both numbers as a product of their prime factors: \(48 = 2^4 \times 3\) and \(60 = 2^2 \times 3 \times 5\). The LCM is found by taking the highest power of each prime factor present: \(LCM = 2^4 \times 3 \times 5 = 16 \times 3 \times 5 = 240\).

M1 for writing prime factorisations of 48 and 60, or listing multiples of both numbers up to at least 120. A1 for 240.

The difference in the ratio parts is \(5 - 3 = 2\) parts. Since 2 parts correspond to 24 counters, 1 part corresponds to \(24 \div 2 = 12\) counters. The total number of parts is \(3 + 5 = 8\) parts. The total number of counters is \(8 \times 12 = 96\).

M1 for finding the value of 1 part by dividing 24 by the difference in parts \((5 - 3)\). A1 for 96.

Since \(y \propto \frac{1}{x^2}\), we have \(y = \frac{k}{x^2}\). Substitute the given values to find \(k\): \(4 = \frac{k}{3^2} \implies 4 = \frac{k}{9} \implies k = 36\). Thus, the equation is \(y = \frac{36}{x^2}\). For \(x = 6\), \(y = \frac{36}{6^2} = \frac{36}{36} = 1\).

M1 for setting up the equation \(y = \frac{k}{x^2}\) and finding \(k = 36\). A1 for 1.

First differences: 7, 11, 15, 19. Second differences: 4, 4, 4. Since the second difference is 4, the coefficient of \(n^2\) is \(4 \div 2 = 2\). Subtracting \(2n^2\) from the original sequence: \(4 - 2(1)^2 = 2\), \(11 - 2(2)^2 = 3\), \(22 - 2(3)^2 = 4\), which gives the linear sequence 2, 3, 4, ... with \(n\)-th term \(n + 1\). Thus, the overall \(n\)-th term is \(2n^2 + n + 1\).

M1 for finding that the second difference is 4 and identifying the \(2n^2\) term, or finding the linear sequence of differences. A1 for \(2n^2 + n + 1\).

Convert the mixed numbers to improper fractions: \(2 \frac{1}{3} = \frac{7}{3}\) and \(1 \frac{2}{5} = \frac{7}{5}\). Now perform the division: \(\frac{7}{3} \div \frac{7}{5} = \frac{7}{3} \times \frac{5}{7} = \frac{5}{3}\). Convert \(\frac{5}{3}\) back into a mixed number: \(1 \frac{2}{3}\).

M1 for writing both mixed numbers as improper fractions, i.e., \(\frac{7}{3}\) and \(\frac{7}{5}\). A1 for \(1 \frac{2}{3}\).

The gradient of the given line is 3. The gradient of the perpendicular line is the negative reciprocal: \(m = -\frac{1}{3}\). Using the equation \(y = mx + c\) with the point \((6, 2)\): \(2 = -\frac{1}{3}(6) + c \implies 2 = -2 + c \implies c = 4\). So, the equation is \(y = -\frac{1}{3}x + 4\).

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

First, take the reciprocal of the fraction to remove the negative exponent: \( \left(\frac{64}{125}\right)^{-\frac{2}{3}} = \left(\frac{125}{64}\right)^{\frac{2}{3}} \). Next, apply the cube root to both the numerator and the denominator: \( \sqrt[3]{\frac{125}{64}} = \frac{5}{4} \). Finally, square the result: \( \left(\frac{5}{4}\right)^2 = \frac{25}{16} \).

M1 for demonstrating a correct step with indices, such as finding the reciprocal \( \left(\frac{125}{64}\right)^{\frac{2}{3}} \) or finding the cube root of the bases to get \( \left(\frac{4}{5}\right)^{-2} \). A1 for \( \frac{25}{16} \) or equivalent fraction or decimal (e.g. \( 1\frac{9}{16} \) or 1.5625).

To solve, first clear the fractions by multiplying every term by the common denominator, 6: \( 3(3x - 1) - 2(x + 2) = 3(7) \). Expand the brackets: \( 9x - 3 - 2x - 4 = 21 \). Simplify the left side by collecting like terms: \( 7x - 7 = 21 \). Add 7 to both sides: \( 7x = 28 \). Divide by 7: \( x = 4 \).

M1 for a correct first step to clear fractions by multiplying all terms by 6 to get \( 3(3x - 1) - 2(x + 2) = 21 \) or equivalent (e.g., expressing left-hand side as a single fraction: \( \frac{3(3x - 1) - 2(x + 2)}{6} = \frac{21}{6} \)). A1 for the correct answer 4.

First, solve the linear inequality:
\(3x - 5 > 2x + 1\)
\(3x - 2x > 1 + 5\)
\(x > 6\)

Next, solve the quadratic inequality:
\(x^2 - 7x - 8 \le 0\)
Factorising the quadratic gives:
\((x - 8)(x + 1) \le 0\)
The critical values are \(x = -1\) and \(x = 8\).
Since we want the expression to be less than or equal to 0, the solution is:
\(-1 \le x \le 8\)

Now, find the values of \(x\) that satisfy both conditions:
\(x > 6\) and \(-1 \le x \le 8\)
Combining these gives:
\(6 < x \le 8\)

The integer values of \(x\) in this range are \(7\) and \(8\).

M1: for solving the first inequality to get \(x > 6\)
M1: for factorising the quadratic expression to find critical values \(x = 8\) and \(x = -1\)
M1: for establishing the combined range \(6 < x \le 8\)
A1.8: for correctly identifying the integers 7 and 8

Let the number of French, German, and Spanish students be \(F\), \(G\), and \(S\) respectively.
We are given:
\(F : G = 5 : 3\)
\(G : S = 4 : 7\)

To find a combined ratio \(F : G : S\), we find a common multiple for \(G\). The lowest common multiple of 3 and 4 is 12.
Scale the first ratio by 4:
\(F : G = 20 : 12\)
Scale the second ratio by 3:
\(G : S = 12 : 21\)

Thus, the combined ratio is:
\(F : G : S = 20 : 12 : 21\)

We are given that \(F = 180\).
Let \(20\text{ parts} = 180\text{ students}\).
\(1\text{ part} = 180 \div 20 = 9\text{ students}\).

The total number of students studying German and Spanish is represented by \(12 + 21 = 33\) parts.
Total students = \(33 \times 9 = 297\).

M1: for attempting to link the two ratios using a common term for German, e.g., \(F:G = 20:12\) and \(G:S = 12:21\)
M1: for writing the combined ratio \(F:G:S = 20:12:21\) or finding the value of 1 part (e.g. \(180 \div 5 = 36\) French groups, German is \(36 \times 3 = 108\))
M1: for calculating German as 108 and Spanish as 189
A1.8: for 297

Since \(y\) is inversely proportional to the square of \((x - 1)\), we can write:
\(y = \frac{k}{(x - 1)^2}\) for some constant \(k\).

Substitute the given values \(x = 3\) and \(y = 9\) to find \(k\):
\(9 = \frac{k}{(3 - 1)^2}\)
\(9 = \frac{k}{2^2}\)
\(9 = \frac{k}{4}\)
\(k = 36\)

So the formula is:
\(y = \frac{36}{(x - 1)^2}\)

Now find \(x\) when \(y = 4\):
\(4 = \frac{36}{(x - 1)^2}\)
\((x - 1)^2 = \frac{36}{4}\)
\((x - 1)^2 = 9\)

Taking the square root of both sides:
\(x - 1 = 3\) or \(x - 1 = -3\)
\(x = 4\) or \(x = -2\)

Since we are looking for the positive value of \(x\), we choose \(x = 4\).

M1: for writing the proportional relationship as \(y = \frac{k}{(x - 1)^2}\)
M1: for substituting \(x = 3, y = 9\) to find \(k = 36\)
M1: for setting up the equation \(4 = \frac{36}{(x - 1)^2}\) and solving to \((x - 1)^2 = 9\)
A1.8: for the positive solution \(x = 4\)

To find the points of intersection, set the two equations equal to each other:
\(2x - 1 = x^2 - 3x + 3\)

Rearrange the equation into standard quadratic form:
\(x^2 - 5x + 4 = 0\)

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

This gives the \(x\)-coordinates of the intersection points:
\(x = 1\) or \(x = 4\)

Now, substitute these \(x\)-values back into the linear equation \(y = 2x - 1\) to find the corresponding \(y\)-coordinates:
For \(x = 1\):
\(y = 2(1) - 1 = 1\)
So, the first intersection point is \((1, 1)\).

For \(x = 4\):
\(y = 2(4) - 1 = 7\)
So, the second intersection point is \((4, 7)\).

The coordinates of the points of intersection are \((1, 1)\) and \((4, 7)\).

M1: for setting the equations equal to each other: \(2x - 1 = x^2 - 3x + 3\)
M1: for rearranging to form the quadratic equation \(x^2 - 5x + 4 = 0\) and attempting to factorise
M1: for finding both correct \(x\)-values: \(x = 1\) and \(x = 4\)
A1.8: for both correct coordinate pairs \((1, 1)\) and \((4, 7)\)

The total number of counters in the bag is \(n + 6\).

The probability of choosing a blue counter on the first draw is:
\(P(\text{First Blue}) = \frac{6}{n + 6}\)

Since the selection is without replacement, there are now \(5\) blue counters left and a total of \(n + 5\) counters remaining.
The probability of choosing a blue counter on the second draw is:
\(P(\text{Second Blue}) = \frac{5}{n + 5}\)

The probability of choosing two blue counters is:
\(P(\text{Both Blue}) = \frac{6}{n + 6} \times \frac{5}{n + 5} = \frac{30}{(n + 6)(n + 5)}\)

We are given that this probability is \(\frac{1}{3}\):
\(\frac{30}{(n + 6)(n + 5)} = \frac{1}{3}\)

Cross-multiply to solve for \(n\):
\(90 = (n + 6)(n + 5)\)
\(90 = n^2 + 11n + 30\)
\(n^2 + 11n - 60 = 0\)

Factorise the quadratic equation:
\((n + 15)(n - 4) = 0\)

This gives \(n = -15\) or \(n = 4\).
Since the number of red counters \(n\) must be a positive integer, we have:
\(n = 4\).

M1: for writing the probability of drawing two blue counters as \(\frac{6}{n + 6} \times \frac{5}{n + 5}\)
M1: for setting up the equation \(\frac{30}{(n+6)(n+5)} = \frac{1}{3}\) and expanding the denominator
M1: for forming the quadratic equation \(n^2 + 11n - 60 = 0\) and attempting to factorise it
A1.8: for selecting the positive root \(n = 4\)

Let \(A\) denote the angle \(BAC\). The area of a triangle is given by the formula:
\(\text{Area} = \frac{1}{2} b c \sin A\)

Substitute the given values into the formula:
\(10\sqrt{3} = \frac{1}{2} \times 5 \times 8 \times \sin A\)
\(10\sqrt{3} = 20 \sin A\)

Solve for \(\sin A\):
\(\sin A = \frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2}\)

Since angle \(BAC\) is acute (less than \(90^\circ\)), we have:
\(A = 60^\circ\)

Now, use the Cosine Rule to calculate the length of side \(BC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC) \cos A\)
\(BC^2 = 5^2 + 8^2 - 2(5)(8) \cos 60^\circ\)

Since \(\cos 60^\circ = \frac{1}{2}\):
\(BC^2 = 25 + 64 - 80 \times \frac{1}{2}\)
\(BC^2 = 89 - 40\)
\(BC^2 = 49\)

Taking the square root:
\(BC = 7\text{ cm}\).

M1: for substituting values into the area formula: \(10\sqrt{3} = \frac{1}{2} \times 5 \times 8 \times \sin A\)
M1: for finding \(\sin A = \frac{\sqrt{3}}{2}\) and identifying that \(A = 60^\circ\)
M1: for applying the Cosine Rule: \(BC^2 = 5^2 + 8^2 - 2(5)(8) \cos 60^\circ\)
A1.8: for finding \(BC = 7\)

From the first equation, we can express \(y\) in terms of \(x\):
\(y = 5 - x\)

Substitute this expression for \(y\) into the second equation:
\(x^2 + (5 - x)^2 = 13\)

Expand and simplify the equation:
\(x^2 + (25 - 10x + x^2) = 13\)
\(2x^2 - 10x + 25 = 13\)
\(2x^2 - 10x + 12 = 0\)

Divide the entire equation by 2 to simplify:
\(x^2 - 5x + 6 = 0\)

Factorise the quadratic:
\((x - 2)(x - 3) = 0\)

This gives two possible values for \(x\):
\(x = 2\) or \(x = 3\)

Now, find the corresponding values of \(y\) using \(y = 5 - x\):
If \(x = 2\):
\(y = 5 - 2 = 3\)

If \(x = 3\):
\(y = 5 - 3 = 2\)

So the solutions are:
\(x = 2, y = 3\) or \(x = 3, y = 2\).

M1: for rearranging the linear equation to get \(y = 5 - x\) or \(x = 5 - y\) and substituting it into the quadratic equation
M1: for expanding and simplifying to get a quadratic equation, e.g. \(2x^2 - 10x + 12 = 0\) or \(x^2 - 5x + 6 = 0\)
M1: for factorising to find \(x = 2\) and \(x = 3\) (or \(y = 2\) and \(y = 3\))
A1.8: for both correct pairs of solutions: \(x = 2, y = 3\) and \(x = 3, y = 2\)

Let the terms of the sequence be represented by:
\(u_1 = 4, u_2 = 11, u_3 = 20, u_4 = 31\)

First, find the first differences between consecutive terms:
\(11 - 4 = 7\)
\(20 - 11 = 9\)
\(31 - 20 = 11\)
The first differences are \(7, 9, 11\).

Now, find the second differences:
\(9 - 7 = 2\)
\(11 - 9 = 2\)
The second difference is constant at \(2\).

Since the second difference is \(2\), the coefficient of the \(n^2\) term is:
\(a = \frac{2}{2} = 1\)

Now, subtract \(n^2\) from each term of the original sequence to find the remaining linear part \(bn + c\):
For \(n = 1\): \(4 - 1^2 = 3\)
For \(n = 2\): \(11 - 2^2 = 7\)
For \(n = 3\): \(20 - 3^2 = 11\)
For \(n = 4\): \(31 - 4^2 = 15\)

The resulting sequence is \(3, 7, 11, 15, \dots\), which is an arithmetic progression.
The common difference is \(4\), so the term is \(4n + c\).
For \(n = 1\), \(4(1) + c = 3 \implies c = -1\).

So the linear part is \(4n - 1\).

Combining the parts, the \(n\)th term of the quadratic sequence is:
\(n^2 + 4n - 1\).

M1: for finding the second differences to be constant at 2
M1: for deducing the coefficient of \(n^2\) is 1, so the sequence contains \(n^2\)
M1: for attempting to subtract \(n^2\) from the sequence terms to get \(3, 7, 11, 15\) and finding the linear rule \(4n - 1\)
A1.8: for the correct expression \(n^2 + 4n - 1\)

Multiply both sides of the equation by \((x+1)(x-2)\) to clear the fractions: \(3x(x-2) - 2(x+1) = (x+1)(x-2)\). Expand the brackets: \(3x^2 - 6x - 2x - 2 = x^2 - x - 2\). Simplify both sides: \(3x^2 - 8x - 2 = x^2 - x - 2\). Subtract \(x^2 - x - 2\) from both sides to form a quadratic equation equal to zero: \(2x^2 - 7x = 0\). Factorise the quadratic expression: \(x(2x - 7) = 0\). Solve for \(x\): \(x = 0\) or \(2x - 7 = 0 \implies x = 3.5\).

M1 for attempting to eliminate fractions by multiplying by \((x+1)(x-2)\). M1 for expanding brackets correctly to obtain \(3x^2 - 8x - 2 = x^2 - x - 2\). M1 for simplifying to a quadratic equation of the form \(2x^2 - 7x = 0\). M1 for factorising the quadratic equation to find at least one correct value of \(x\). A1 for both solutions: \(x = 0\) and \(x = 3.5\) (or \(\frac{7}{2}\)).

The probability of choosing two lemon sweets is given by: \(\frac{6}{n} \times \frac{5}{n-1} = \frac{1}{3}\). Simplifying this gives: \(\frac{30}{n(n-1)} = \frac{1}{3} \implies n(n-1) = 90 \implies n^2 - n - 90 = 0\). Factorising the quadratic equation: \((n - 10)(n + 9) = 0\). Since the number of sweets \(n\) must be positive, \(n = 10\). This means there are 10 sweets in total: 6 lemon sweets and \(10 - 6 = 4\) orange sweets. The probability of choosing two sweets of different flavours (Lemon then Orange, or Orange then Lemon) is: \(P(\text{Lemon, Orange}) + P(\text{Orange, Lemon}) = \left(\frac{6}{10} \times \frac{4}{9}\right) + \left(\frac{4}{10} \times \frac{6}{9}\right) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}\).

M1 for setting up the equation for the probability of choosing two lemon sweets: \(\frac{6}{n} \times \frac{5}{n-1} = \frac{1}{3}\). M1 for rearranging to form the quadratic equation \(n^2 - n - 90 = 0\) and finding the valid solution \(n = 10\). M1 for deducing there are 4 orange sweets. M1 for a complete method to calculate the probability of different flavours: \(\frac{6}{10} \times \frac{4}{9} + \frac{4}{10} \times \frac{6}{9}\). A1 for the final answer of \(\frac{8}{15}\) (or equivalent fraction).

Since \(y\) is inversely proportional to the square of \(x\), we can write: \(y = \frac{k}{x^2}\) where \(k\) is a constant. Using the first set of values: \(16 = \frac{k}{r^2} \implies k = 16r^2\). Using the second set of values: \(9 = \frac{k}{(r+1)^2} \implies k = 9(r+1)^2\). Equating the expressions for \(k\): \(16r^2 = 9(r+1)^2\). Taking the positive square root of both sides (since \(r > 0\)): \(4r = 3(r+1) \implies 4r = 3r + 3 \implies r = 3\). Now find the constant \(k\): \(k = 16 \times 3^2 = 144\). When \(x = 2r\), we have \(x = 2 \times 3 = 6\). Find the value of \(y\): \(y = \frac{144}{6^2} = \frac{144}{36} = 4\).

M1 for establishing the relationship \(y = \frac{k}{x^2}\). M1 for forming the equation \(16r^2 = 9(r+1)^2\). M1 for solving this equation to find \(r = 3\). M1 for finding the value of the constant \(k = 144\) and substituting \(x = 6\) into the formula. A1 for the final answer of \(4\).

The area of a triangle is given by \(\frac{1}{2} a c \sin B\). Here, \(\text{Area} = \frac{1}{2} \times x \times (x+3) \times \sin 60^\circ\). Since \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), we have: \(10\sqrt{3} = \frac{1}{2} \times x(x+3) \times \frac{\sqrt{3}}{2} \implies 10\sqrt{3} = \frac{\sqrt{3}}{4} x(x+3)\). Dividing both sides by \(\sqrt{3}\): \(10 = \frac{1}{4} x(x+3) \implies 40 = x(x+3) \implies x^2 + 3x - 40 = 0\). Factorising the quadratic equation: \((x + 8)(x - 5) = 0\). Since length must be positive, \(x = 5\). Thus, the side lengths are \(AB = 5\) cm and \(BC = 5 + 3 = 8\) cm. Using the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos 60^\circ\). Since \(\cos 60^\circ = 0.5\): \(AC^2 = 5^2 + 8^2 - 2 \times 5 \times 8 \times 0.5 = 25 + 64 - 40 = 49\). Taking the square root: \(AC = \sqrt{49} = 7\) cm.

M1 for writing a correct equation for the area of the triangle: \(\frac{1}{2} x (x+3) \sin 60^\circ = 10\sqrt{3}\). M1 for simplifying this to the quadratic equation \(x^2 + 3x - 40 = 0\) and solving to find \(x = 5\). M1 for finding the lengths of \(AB\) (5 cm) and \(BC\) (8 cm). M1 for substituting these side lengths and \(\cos 60^\circ\) into the Cosine Rule formula. A1 for the final answer of \(7\) (or 7 cm).

To solve the inequality \(2x^2 + 5x - 3 \le 0\), we first find the critical values by factorising the quadratic expression:
\(2x^2 + 5x - 3 = 0\)
\((2x - 1)(x + 3) = 0\)

The critical values are \(x = 0.5\) and \(x = -3\).
Since the inequality is \(\le 0\), the solution range is:
\(-3 \le x \le 0.5\)

The integer values of \(x\) that fall within this range are \(-3, -2, -1, 0\).

M1: Attempt to factorise the quadratic or use the quadratic formula to find the critical values of \(-3\) and \(0.5\).
A1: Correct inequality range identified as \(-3 \le x \le 0.5\).
A0.5: All four correct integer values listed: \(-3, -2, -1, 0\) (order does not matter).

The duration from the start of 2020 to the start of 2025 is exactly \(5\) years.
The decimal multiplier for an increase of \(8\%\) is \(1.08\).
Using the compound growth formula:
\(\text{Value} = 1500 \times 1.08^5\)
\(\text{Value} = 1500 \times 1.469328077...\)
\(\text{Value} = 2203.992115...\)

To the nearest penny (2 decimal places), this is \(\pounds 2203.99\).

M1: Identifies compound interest multiplier of \(1.08\) and raises to the power of \(5\) (e.g. \(1500 \times 1.08^5\)).
A1: Correct calculation giving \(2203.99...\)
A0.5: Rounds correctly to the nearest penny (2 decimal places) giving \(2203.99\).

The area of a triangle can be calculated using the formula:
\(\text{Area} = \frac{1}{2} a c \sin B\)

Substitute the given values into the formula:
\(\text{Area} = \frac{1}{2} \times 9.5 \times 7.2 \times \sin(43^\circ)\)
\(\text{Area} = 34.2 \times 0.68199836...\)
\(\text{Area} \approx 23.3243...\text{ cm}^2\)

Rounding to 3 significant figures gives \(23.3\text{ cm}^2\).

M1: Correct substitution of values into the area formula \(\frac{1}{2} a b \sin C\) (e.g. \(0.5 \times 7.2 \times 9.5 \times \sin(43)\)).
A1: Evaluation to a minimum of 3 decimal places (\(23.324...\)).
A0.5: Correct rounding to 3 significant figures (\(23.3\)).

Let's find the first and second differences:
Terms: \(4, \quad 11, \quad 22, \quad 37, \quad 56\)
First differences: \(7, \quad 11, \quad 15, \quad 19\)
Second differences: \(4, \quad 4, \quad 4\)

Since the second difference is constant at \(4\), the coefficient of the \(n^2\) term is \(4 \div 2 = 2\).

Now, we subtract \(2n^2\) from the original terms of the 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 \(n=4\): \(37 - 2(4)^2 = 5\)
For \(n=5\): \(56 - 2(5)^2 = 6\)

The linear sequence of differences is \(2, 3, 4, 5, 6, \dots\), which has the formula \(n + 1\).

Combining the two parts gives the \(n\)-th term of the quadratic sequence:
\(2n^2 + n + 1\).

M1: Identifies that the second difference is \(4\) and correctly finds the leading coefficient of \(2n^2\).
A1: Subtracts \(2n^2\) to find the remaining linear sequence \(n + 1\) (or solves for linear coefficients using simultaneous equations).
A0.5: Combines terms to state the final expression \(2n^2 + n + 1\).

Since \(P\) is inversely proportional to the square root of \(w\), we write:
\(P = \frac{k}{\sqrt{w}}\)

Using the given values \(w = 16\) and \(P = 15\):
\(15 = \frac{k}{\sqrt{16}}\)
\(15 = \frac{k}{4} \implies k = 60\)

So the formula relating \(P\) and \(w\) is:
\(P = \frac{60}{\sqrt{w}}\)

Now, substitute \(w = 25\) into the formula:
\(P = \frac{60}{\sqrt{25}} = \frac{60}{5} = 12\).

M1: Sets up the proportional relationship equation \(P = \frac{k}{\sqrt{w}}\) and substitutes the known values.
A1: Correctly calculates the constant of proportionality \(k = 60\).
A0.5: Calculates the correct final value of \(P = 12\).

To find the upper bound for the density, we need the upper bound of the mass divided by the lower bound of the volume:
\(\text{Density}_{\text{UB}} = \frac{\text{Mass}_{\text{UB}}}{\text{Volume}_{\text{LB}}}\)

1. Mass is \(425\text{ g}\) to the nearest \(5\text{ g}\):
\(\text{Mass}_{\text{UB}} = 425 + 2.5 = 427.5\text{ g}\)

2. Volume is \(50\text{ cm}^3\) to the nearest \(2\text{ cm}^3\):
\(\text{Volume}_{\text{LB}} = 50 - 1 = 49\text{ cm}^3\)

3. Calculate the upper bound of density:
\(\text{Density}_{\text{UB}} = \frac{427.5}{49} \approx 8.724489...\text{ g/cm}^3\)

Rounding to 2 decimal places gives \(8.72\text{ g/cm}^3\).

M1: Identifies either the correct upper bound of mass (\(427.5\)) or the correct lower bound of volume (\(49\)).
A1: Correct division calculation of \(\frac{427.5}{49}\).
A0.5: Rounds the final answer correctly to 2 decimal places to get \(8.72\).

To calculate the composite function value \(fg(3)\), we evaluate the inner function \(g(3)\) first:
\(g(3) = 3^2 + 2 = 9 + 2 = 11\)

Now, substitute this result into the outer function \(f(x)\):
\(f(11) = 3(11) - 5 = 33 - 5 = 28\).

M1: Correctly evaluates \(g(3) = 11\) or expresses the composite function algebraically as \(fg(x) = 3(x^2 + 2) - 5\).
A1: Show substitution of the value into the final step: \(3(11) - 5\).
A0.5: Obtains the correct integer answer of \(28\).

The total surface area of a hemisphere consists of the flat circular base area (\(\pi r^2\)) and the curved surface area (\(2\pi r^2\)):
\(\text{Total Surface Area} = \pi r^2 + 2\pi r^2 = 3\pi r^2\)

We are given the total surface area is \(108\pi\text{ cm}^2\):
\(3\pi r^2 = 108\pi\)

Divide both sides by \(\pi\):
\(3r^2 = 108\)

Divide by \(3\):
\(r^2 = 36\)

Take the positive square root:
\(r = 6\text{ cm}\).

M1: Correctly identifies the formula for the total surface area of a hemisphere as \(3\pi r^2\) and equates it to \(108\pi\).
A1: Simplifies equation correctly to obtain \(r^2 = 36\).
A0.5: Identifies the correct radius value \(6\).

Since \(y\) is inversely proportional to the cube root of \(x\), we can write the relationship as: \(y = \frac{k}{\sqrt[3]{x}}\) where \(k\) is a constant. Substitute the given values \(x = 64\) and \(y = 3\) to find \(k\): \(3 = \frac{k}{\sqrt[3]{64}}\) Since \(\sqrt[3]{64} = 4\), we have: \(3 = \frac{k}{4} \implies k = 12\). Now, substitute \(k = 12\) and \(y = 2.4\) into the equation to find \(x\): \(2.4 = \frac{12}{\sqrt[3]{x}}\) Rearrange to solve for \(\sqrt[3]{x}\): \(\sqrt[3]{x} = \frac{12}{2.4} = 5\). Cube both sides to find \(x\): \(x = 5^3 = 125\).

M1: Set up the correct proportional equation and find the constant \(k = 12\) (or equivalent step showing \(y \sqrt[3]{x} = 12\)). M1: Substitute \(y = 2.4\) to obtain \(\sqrt[3]{x} = 5\) (or equivalent algebraic step). A0.5: Correct final answer of 125.

The area of a triangle can be calculated using the formula: \(\text{Area} = \frac{1}{2} a b \sin(C)\). Let \(\theta\) represent angle \(BAC\). Substituting the known values: \(54 = \frac{1}{2} \times 12 \times 12 \times \sin(\theta)\) \(54 = 72 \sin(\theta)\) \(\sin(\theta) = \frac{54}{72} = 0.75\). Finding the acute angle: \(\theta_{\text{acute}} = \sin^{-1}(0.75) \approx 48.59^\circ\). Since angle \(BAC\) is given as obtuse, we find: \(\theta = 180^\circ - 48.59^\circ = 131.41^\circ\). Rounding to 1 decimal place gives \(131.4^\circ\).

M1: Substitute correct values into the sine area formula to get \(54 = 0.5 \times 12 \times 12 \times \sin(\theta)\) or find \(\sin(\theta) = 0.75\). M1: Obtain the acute angle \(\approx 48.6^\circ\) or state the method for finding the obtuse angle using \(180^\circ - \text{acute angle}\). A0.5: Correct final answer of 131.4 (accept 131.4 or 131.41).

Let \(r\) be the multiplier for the annual increase. After 2 years, from the start of 2018 to the start of 2020, the value is \(40000 \times r^2 = 46225\). Dividing both sides by 40,000 gives \(r^2 = 1.155625\). Taking the positive square root gives \(r = 1.075\), representing a 7.5% annual increase. To find the value at the start of 2023, which is 5 years after the start of 2018, we calculate \(40000 \times 1.075^5 = 57425.1734...\). To the nearest pound, the value is £57,425.

M1: For setting up the equation for the multiplier, e.g., \(40000r^2 = 46225\). M1: For solving to find \(r = 1.075\). M1: For calculating the value after 5 years, e.g., \(40000 \times 1.075^5\) or \(46225 \times 1.075^3\). A1.7: For the correct final answer of 57425 (accept 57425.17 or 57425).

We can use the Cosine Rule to find the side \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\). Substituting the given values: \(AC^2 = 75^2 + 110^2 - 2 \times 75 \times 110 \times \cos(58^\circ)\). This simplifies to \(AC^2 = 5625 + 12100 - 16500 \times \cos(58^\circ)\). Using a calculator: \(AC^2 = 17725 - 8743.6678... = 8981.3321...\). Taking the square root gives \(AC = 94.7698...\text{ m}\). Correct to 1 decimal place, the length is 94.8 m.

M1: For substituting correctly into the Cosine Rule, i.e., \(75^2 + 110^2 - 2(75)(110)\cos(58^\circ)\). A1: For evaluating intermediate terms to get \(17725 - 16500\cos(58^\circ)\) or \(8981.3\). M1: For attempting to find the square root of their intermediate value. A1.7: For the final answer of 94.8.

Substitute the linear equation \(y = 2x - 4\) into the quadratic equation: \(2x^2 - (2x - 4)^2 = 14\). Expand the squared term: \(2x^2 - (4x^2 - 16x + 16) = 14\). Simplify to get: \(-2x^2 + 16x - 16 = 14\). Rearrange to form a quadratic equation set to zero: \(-2x^2 + 16x - 30 = 0\). Divide all terms by -2 to simplify: \(x^2 - 8x + 15 = 0\). Factorising gives: \((x - 3)(x - 5) = 0\). Thus, the x-values are \(x = 3\) or \(x = 5\). Substitute these back into \(y = 2x - 4\): when \(x = 3\), \(y = 2(3) - 4 = 2\); when \(x = 5\), \(y = 2(5) - 4 = 6\). This yields the solutions (3, 2) and (5, 6).

M1: For substituting \(y = 2x - 4\) into the first equation. A1: For expanding and simplifying to a three-term quadratic equation, e.g., \(x^2 - 8x + 15 = 0\) or \(-2x^2 + 16x - 30 = 0\). M1: For solving their quadratic equation to find two values of \(x\). A1.7: For both correct pairs (3, 2) and (5, 6) in the correct format.

The probability of choosing two orange sweets without replacement is given by: \(\frac{6}{n} \times \frac{5}{n-1} = \frac{1}{3}\). This simplifies to \(\frac{30}{n(n-1)} = \frac{1}{3}\). Cross-multiplying gives \(n(n-1) = 90\), which expands and rearranges to the quadratic equation \(n^2 - n - 90 = 0\). Factorising this gives \((n - 10)(n + 9) = 0\). Since \(n\) must be a positive integer, \(n = 10\). This represents the total number of sweets in the bag. The number of green sweets is the total number minus the orange sweets: \(10 - 6 = 4\).

M1: For setting up the probability product expression: \(\frac{6}{n} \times \frac{5}{n-1}\). M1: For equating to \(\frac{1}{3}\) and deriving the quadratic equation \(n^2 - n - 90 = 0\). M1: For solving the quadratic equation to find \(n = 10\) (rejecting \(n = -9\)). A1.7: For correctly subtracting 6 from 10 to obtain 4 green sweets.

The surface area of a sphere is given by \(4\pi r^2\). Setting this equal to the given area: \(4\pi r^2 = 144\pi\), which simplifies to \(r^2 = 36\), so \(r = 6\text{ cm}\). The volume of a sphere is \(V = \frac{4}{3}\pi r^3\). Substituting \(r = 6\): \(V = \frac{4}{3}\pi (6^3) = \frac{4}{3}\pi (216) = 288\pi\text{ cm}^3\). The volume of a cone is \(V = \frac{1}{3}\pi r^2 h\). Since the volume of the cone equals the volume of the sphere: \(\frac{1}{3}\pi (6^2) h = 288\pi\). This simplifies to \(12\pi h = 288\pi\). Dividing both sides by \(12\pi\) gives \(h = 24\text{ cm}\).

M1: For equating \(4\pi r^2 = 144\pi\) and solving to find \(r = 6\). M1: For calculating the volume of the sphere as \(288\pi\) or using equivalent algebra. M1: For equating the volume of the cone to their volume of the sphere: \(\frac{1}{3}\pi (6^2) h = 288\pi\). A1.7: For finding the correct height of 24.

First, write the proportional relationship between \(A\) and \(B\): \(A = \frac{k_1}{B^2}\). Substituting \(B = 3\) and \(A = 10\) gives \(10 = \frac{k_1}{3^2}\), so \(k_1 = 90\) and \(A = \frac{90}{B^2}\). Next, write the proportional relationship between \(B\) and \(C\): \(B = k_2 \sqrt{C}\). Substituting \(C = 16\) and \(B = 8\) gives \(8 = k_2 \sqrt{16}\), which means \(8 = 4 k_2\), so \(k_2 = 2\) and \(B = 2\sqrt{C}\). Now, find \(B\) when \(C = 9\): \(B = 2\sqrt{9} = 2 \times 3 = 6\). Finally, substitute \(B = 6\) into the equation for \(A\): \(A = \frac{90}{6^2} = \frac{90}{36} = 2.5\).

M1: For setting up \(A = \frac{k_1}{B^2}\) and finding \(k_1 = 90\). M1: For setting up \(B = k_2\sqrt{C}\) and finding \(k_2 = 2\). M1: For substituting \(C = 9\) to find \(B = 6\). A1.7: For substituting \(B = 6\) to find the final value of 2.5 (accept 5/2).

First, find the critical values by solving the quadratic equation \(3x^2 - 10x - 8 = 0\). Factorising the quadratic gives \(3x^2 - 12x + 2x - 8 = 0\), which groups into \(3x(x - 4) + 2(x - 4) = 0\), and factorises to \((3x + 2)(x - 4) = 0\). The critical values are \(x = -\frac{2}{3}\) and \(x = 4\). Since the inequality is less than 0, the solution is the region between these values: \(-\frac{2}{3} < x < 4\). The integers in this interval are 0, 1, 2, and 3. Therefore, there are 4 integers that satisfy the inequality.

M1: For attempt to factorise or solve \(3x^2 - 10x - 8 = 0\). A1: For finding correct critical values of \(-\frac{2}{3}\) (or \(-0.67\)) and \(4\). M1: For establishing the interval of interest: \(-\frac{2}{3} < x < 4\). A1.7: For correctly identifying the integers as 0, 1, 2, 3 and giving the count of 4.

First, we look at the differences between consecutive terms: 11 - 3 = 8, 23 - 11 = 12, 39 - 23 = 16. The first differences are 8, 12, 16. The second differences are constant: 12 - 8 = 4, 16 - 12 = 4. The coefficient of the \(n^2\) term is half of the second difference, so \(a = 4 / 2 = 2\). Subtracting \(2n^2\) from each term of the sequence gives: Term 1: \(3 - 2(1^2) = 1\), Term 2: \(11 - 2(2^2) = 3\), Term 3: \(23 - 2(3^2) = 5\), Term 4: \(39 - 2(4^2) = 7\). This yields the linear sequence 1, 3, 5, 7, which has the general formula \(2n - 1\). Therefore, the overall general formula for the sequence is \(T_n = 2n^2 + 2n - 1\). To find the 20th term, we substitute \(n = 20\): \(T_{20} = 2(20^2) + 2(20) - 1 = 2(400) + 40 - 1 = 800 + 40 - 1 = 839\).

M1: For finding the constant second difference of 4 and setting \(a = 2\). M1: For subtracting \(2n^2\) from the terms and finding the linear sequence 1, 3, 5, 7. A1: For deriving the correct general term expression \(2n^2 + 2n - 1\). A1.7: For calculating the correct 20th term of 839.

Let the multiplier for the first year be \(1 + \frac{x}{100}\) and for the second year be \(1 + \frac{2x}{100}\). The value at the end of the second year is given by \(5000 \left(1 + \frac{x}{100}\right)\left(1 + \frac{2x}{100}\right) = 5616\). Dividing both sides by 5000 gives \(\left(1 + \frac{x}{100}\right)\left(1 + \frac{2x}{100}\right) = 1.1232\). Expanding the brackets yields \(1 + \frac{3x}{100} + \frac{2x^2}{10000} = 1.1232\). Multiplying the entire equation by 10000 to eliminate the denominators gives \(10000 + 300x + 2x^2 = 11232\). Rearranging into a standard quadratic equation gives \(2x^2 + 300x - 1232 = 0\). Dividing by 2 simplifies this to \(x^2 + 150x - 616 = 0\). Factorising the quadratic equation gives \((x - 4)(x + 154) = 0\). Since \(x\) must be positive, we discard \(x = -154\). Thus, \(x = 4\).

M1: For setting up the initial equation \(5000(1 + \frac{x}{100})(1 + \frac{2x}{100}) = 5616\) or equivalent. M1: For expanding the equation and simplifying to a three-term quadratic equation, e.g., \(2x^2 + 300x - 1232 = 0\) or \(x^2 + 150x - 616 = 0\). M1: For a valid method to solve their quadratic equation, e.g., factorising to \((x - 4)(x + 154) = 0\) or applying the quadratic formula. A1: For the final answer of 4 (accept 4, reject -154).

Using the formula for the area of a triangle, \(\text{Area} = \frac{1}{2} ac \sin B\), we substitute the given values to find the angle: \(48.5 = \frac{1}{2} \times 9.7 \times 12.4 \times \sin(\angle ABC)\), which simplifies to \(48.5 = 60.14 \sin(\angle ABC)\). This gives \(\sin(\angle ABC) = \frac{48.5}{60.14} \approx 0.8064516\). Since angle \(ABC\) is obtuse, we find its value by subtracting the acute angle from \(180^\circ\): \(\angle ABC = 180^\circ - \sin^{-1}(0.8064516) \approx 180^\circ - 53.750^\circ = 126.250^\circ\). Now, we apply the Cosine Rule to find the length \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\). Substituting the values gives: \(AC^2 = 12.4^2 + 9.7^2 - 2 \times 12.4 \times 9.7 \times \cos(126.250^\circ)\). This simplifies to \(AC^2 = 153.76 + 94.09 - 240.56 \times (-0.591312) = 247.85 + 142.246 = 390.096\). Taking the square root gives \(AC = \sqrt{390.096} \approx 19.7508\text{ cm}\). Rounding to 3 significant figures, we get \(19.8\text{ cm}\).

M1: For substituting the values into the area formula: \(\frac{1}{2} \times 12.4 \times 9.7 \times \sin B = 48.5\). M1: For identifying that the angle is obtuse and calculating \(\angle ABC \approx 126.25^\circ\) (or \(126.3^\circ\)). M1: For substituting their angle and side lengths into the Cosine Rule: \(AC^2 = 12.4^2 + 9.7^2 - 2 \times 12.4 \times 9.7 \times \cos(126.25^\circ)\). A1: For the final answer of 19.8 (accept answers in the range 19.7 to 19.8).

The total number of marbles in the box is \(x + 3\). The probability of choosing two red marbles is \(P(\text{Red, Red}) = \frac{x}{x+3} \times \frac{x-1}{x+2}\). The probability of choosing two blue marbles is \(P(\text{Blue, Blue}) = \frac{3}{x+3} \times \frac{2}{x+2} = \frac{6}{(x+3)(x+2)}\). Since the marbles must be of the same colour, the sum of these probabilities is \(\frac{1}{2}\): \(\frac{x(x-1) + 6}{(x+3)(x+2)} = \frac{1}{2}\). Expanding the numerator and denominator gives \(\frac{x^2 - x + 6}{x^2 + 5x + 6} = \frac{1}{2}\). Cross-multiplying gives \(2(x^2 - x + 6) = x^2 + 5x + 6\), which simplifies to \(2x^2 - 2x + 12 = x^2 + 5x + 6\). Rearranging this into standard quadratic form gives \(x^2 - 7x + 6 = 0\). Factorising this quadratic yields \((x-6)(x-1) = 0\). Since \(x > 1\), we reject \(x = 1\) and find that \(x = 6\).

M1: For writing correct probability expressions for both identical pairs in terms of \(x\), e.g., \(\frac{x(x-1)}{(x+3)(x+2)}\) and \(\frac{6}{(x+3)(x+2)}\). M1: For setting up the algebraic equation: \(\frac{x^2 - x + 6}{x^2 + 5x + 6} = \frac{1}{2}\) or equivalent. M1: For simplifying the equation to the quadratic form \(x^2 - 7x + 6 = 0\). A1: For the final answer of 6 (the value \(x = 1\) must be rejected or not chosen).

First, express \(x\) in terms of \(z\) using the direct proportion relationship: \(x = k\sqrt{z}\). Substituting the values \(z = 16\) and \(x = 12\) gives \(12 = k\sqrt{16}\), which means \(12 = 4k\) and thus \(k = 3\). So, \(x = 3\sqrt{z}\). Next, express \(y\) in terms of \(x\) using the inverse proportion relationship: \(y = \frac{c}{x^2}\). Substituting the values \(x = 3\) and \(y = 8\) gives \(8 = \frac{c}{3^2}\), which means \(8 = \frac{c}{9}\) and thus \(c = 72\). So, \(y = \frac{72}{x^2}\). To find a formula for \(y\) in terms of \(z\), we substitute \(x = 3\sqrt{z}\) into the equation for \(y\): \(y = \frac{72}{(3\sqrt{z})^2} = \frac{72}{9z} = \frac{8}{z}\). Finally, to find the value of \(y\) when \(z = 9\), substitute this value into the formula: \(y = \frac{8}{9}\).

M1: For finding the constant of proportionality for \(x\) to get \(x = 3\sqrt{z}\). M1: For finding the constant of proportionality for \(y\) to get \(y = \frac{72}{x^2}\). M1: For substituting the expression for \(x\) into the expression for \(y\) to obtain the simplified formula \(y = \frac{8}{z}\). A1: For the final answer of 8/9 (accept 8/9, reject decimal approximations).