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Convert the mixed number to an improper fraction:
\( 1\frac{3}{5} = \frac{1 \times 5 + 3}{5} = \frac{8}{5} \)

Now divide by \( \frac{2}{3} \) by multiplying by its reciprocal:
\( \frac{8}{5} \div \frac{2}{3} = \frac{8}{5} \times \frac{3}{2} \)

Multiply the numerators and denominators:
\( \frac{8 \times 3}{5 \times 2} = \frac{24}{10} \)

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

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

M1: For converting \( 1\frac{3}{5} \) to \( \frac{8}{5} \) and showing an intention to multiply by the reciprocal \( \frac{3}{2} \) (or equivalent division method using a common denominator, e.g., \( \frac{24}{15} \div \frac{10}{15} \)).
A1: For the correct mixed number in its simplest form: \( 2\frac{2}{5} \).

The green counters correspond to 4 parts of the ratio.

First, find the value of 1 part:
\( 24 \div 4 = 6 \) counters.

The total number of parts in the ratio is:
\( 3 + 5 + 4 = 12 \) parts.

Now find the total number of counters:
\( 12 \times 6 = 72 \).

M1: For a correct step to find the value of one part (e.g., \( 24 \div 4 = 6 \)) or finding the total number of parts (e.g., \( 3 + 5 + 4 = 12 \)).
A1: For 72.

First expand both single brackets:
\( 3(2x - 5) = 6x - 15 \)
\( -2(x - 4) = -2x + 8 \)

Now collect the like terms:
\( 6x - 2x - 15 + 8 \)
\( = 4x - 7 \)

M1: For expanding at least one bracket correctly, resulting in either \( 6x - 15 \) or \( -2x + 8 \).
A1: For the fully simplified expression \( 4x - 7 \).

The sum of the probabilities of all possible outcomes is 1.

Sum the given probabilities:
\( 0.15 + 0.42 + 0.23 = 0.8 \)

Subtract this sum from 1:
\( 1 - 0.8 = 0.2 \)

Therefore, the probability that the spinner lands on 3 is 0.2.

M1: For summing the given probabilities: \( 0.15 + 0.42 + 0.23 \) (or 0.8), or showing the expression \( 1 - 0.15 - 0.42 - 0.23 \).
A1: For 0.2 (or equivalent fraction, e.g., \( \frac{1}{5} \)).

The sum of the exterior angles of any regular polygon is \( 360^\circ \).

Since the polygon is regular, all its exterior angles are equal.

\( \text{Number of sides} = \frac{360^\circ}{\text{Exterior angle}} \)
\( \text{Number of sides} = \frac{360}{40} = 9 \)

M1: For \( 360 \div 40 \) or for finding the interior angle \( 180 - 40 = 140 \) and setting up a correct equation, e.g., \( \frac{(n - 2) \times 180}{n} = 140 \).
A1: For 9.

Find the total number of goals scored by multiplying the number of goals by their corresponding frequency and summing them:
\( (0 \times 2) + (1 \times 4) + (2 \times 3) + (3 \times 1) \)
\( = 0 + 4 + 6 + 3 = 13 \) goals.

The total number of matches is the sum of the frequencies:
\( 2 + 4 + 3 + 1 = 10 \) matches.

Calculate the mean:
\( \text{Mean} = \frac{\text{Total goals}}{\text{Total matches}} = \frac{13}{10} = 1.3 \)

M1: For a correct method to find the total number of goals (e.g., \( 0 \times 2 + 1 \times 4 + 2 \times 3 + 3 \times 1 \) or showing 13), or for dividing their total by 10.
A1: For 1.3 (or \( \frac{13}{10} \) or \( 1\frac{3}{10} \)).

Multiply the numerical coefficients and the powers of 10 separately:
\( (3 \times 4) \times (10^5 \times 10^{-2}) \)
\( = 12 \times 10^{5 + (-2)} \)
\( = 12 \times 10^3 \)

Now, convert \( 12 \times 10^3 \) into standard form:
\( 12 = 1.2 \times 10^1 \)
\( 1.2 \times 10^1 \times 10^3 = 1.2 \times 10^4 \)

M1: For finding \( 12 \times 10^3 \) or 12000, or for an answer of the form \( 1.2 \times 10^k \) where \( k \) is an integer other than 4.
A1: For \( 1.2 \times 10^4 \).

Identify the highest common factor of both terms \( 3x^2 \) and \( -12x \).

The highest common factor of the coefficients 3 and 12 is 3.
The highest common factor of \( x^2 \) and \( x \) is \( x \).

Therefore, the highest common factor of the two terms is \( 3x \).

Divide each term by \( 3x \) to find the terms inside the brackets:
\( \frac{3x^2}{3x} = x \)
\( \frac{-12x}{3x} = -4 \)

Thus, the fully factorised expression is:
\( 3x(x - 4) \)

M1: For identifying a common factor and factorising at least partially, e.g., \( 3(x^2 - 4x) \) or \( x(3x - 12) \).
A1: For the fully factorised expression \( 3x(x - 4) \).

First, convert the mixed numbers to improper fractions:
\( 3 \frac{3}{4} = \frac{15}{4} \) and \( 1 \frac{1}{8} = \frac{9}{8} \).

Now, divide the fractions:
\( \frac{15}{4} \div \frac{9}{8} = \frac{15}{4} \times \frac{8}{9} \).

Simplify by cancelling:
\( \frac{15 \times 8}{4 \times 9} = \frac{5 \times 2}{1 \times 3} = \frac{10}{3} \).

Convert back to a mixed number:
\( \frac{10}{3} = 3 \frac{1}{3} \).

M1: For converting at least one mixed number to a correct improper fraction (e.g., \( \frac{15}{4} \) or \( \frac{9}{8} \)) OR showing multiplication by the reciprocal of the second fraction.
A1: For the correct mixed number \( 3 \frac{1}{3} \) (accept \( \frac{10}{3} \) for method, but final mark requires the simplified mixed number).

Collect the like terms:
For \( x^2 \): \( 5x^2 + 2x^2 = 7x^2 \).
For \( x \): \( -3x + 7x = 4x \).

Combining these gives:
\( 7x^2 + 4x \).

M1: For correctly simplifying either the \( x^2 \) term to \( 7x^2 \) or the \( x \) term to \( 4x \).
A1: For \( 7x^2 + 4x \) or equivalent.

Calculate the actual distance in centimetres:
\( 6\text{ cm} \times 25000 = 150000\text{ cm} \).

Convert centimetres to metres by dividing by 100:
\( 150000 \div 100 = 1500\text{ m} \).

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

M1: For \( 6 \times 25000 \) or \( 150000 \) seen, or for a correct conversion process from cm to km (e.g. dividing by \( 100000 \)).
A1: For \( 1.5 \) (accept \( 1.5\text{ km} \)).

The formula for the circumference of a circle is:
\( C = \pi d \)

Substitute \( d = 14 \) and \( \pi = \frac{22}{7} \):
\( C = \frac{22}{7} \times 14 \)
\( C = 22 \times 2 = 44\text{ cm} \).

M1: For a correct substitution into the circumference formula, e.g., \( \frac{22}{7} \times 14 \) or \( 2 \times \frac{22}{7} \times 7 \).
A1: For \( 44 \).

The sum of all probabilities in a single event is 1.
\( P(\text{green}) = 1 - (P(\text{red}) + P(\text{blue})) \)
\( P(\text{green}) = 1 - (0.25 + 0.4) = 1 - 0.65 = 0.35 \).

M1: For \( 0.25 + 0.4 \) or \( 0.65 \), or \( 1 - 0.25 - 0.4 \).
A1: For \( 0.35 \) (or equivalent fraction, e.g., \( \frac{7}{20} \) or percentage \( 35\% \)).

First, find the total sum of the rents:
\( 180 + 220 + 190 + 250 + 210 = 1050 \).

Now, divide the total sum by the number of flats (5):
\( 1050 \div 5 = 210 \).

M1: For showing the sum of the numbers divided by 5, e.g., \( \frac{1050}{5} \) or \( (180 + 220 + 190 + 250 + 210) \div 5 \).
A1: For \( 210 \) (or \( £210 \)).

First, address the negative index by taking the reciprocal of the fraction:
\( \left(\frac{27}{64}\right)^{-\frac{2}{3}} = \left(\frac{64}{27}\right)^{\frac{2}{3}} \).

Next, apply the cube root (denominator of the fraction exponent):
\( \left(\frac{64}{27}\right)^{\frac{1}{3}} = \frac{\sqrt[3]{64}}{\sqrt[3]{27}} = \frac{4}{3} \).

Finally, square the result (numerator of the fraction exponent):
\( \left(\frac{4}{3}\right)^2 = \frac{16}{9} \).

M1: For resolving the negative index to \( \left(\frac{64}{27}\right)^{\frac{2}{3}} \) OR applying the cube root to get \( \left(\frac{3}{4}\right)^{-2} \) or \( \frac{9}{16} \) at any stage.
A1: For \( \frac{16}{9} \) or \( 1 \frac{7}{9} \).

Let \(s\) be the number of small boxes and \(l\) be the number of large boxes. The total number of pencils is \(12s + 30l = 240\), which simplifies to \(2s + 5l = 40\). Solving for \(s\) gives \(s = 20 - 2.5l\). Let \(x\) be the cost of a small box and \(y\) be the cost of a large box. We are given \(5x = 2y\), which means \(y = 2.5x\). The total cost is \(s \times x + l \times y = (20 - 2.5l) \times x + l \times (2.5x) = 20x - 2.5lx + 2.5lx = 20x\). We are given the total cost is %48, so \(20x = 48\). This gives \(x = 48 / 20 = 2.4\). Therefore, the cost of a single small box of pencils is %2.40.

M1: for setting up the equation for the number of pencils \(12s + 30l = 240\) or writing the relationship between the costs \(5x = 2y\). M1: for expressing the total cost in terms of a single variable, showing that the total cost is \(20x\). A1: for %2.40 (or 2.4).

The volume of the metal cuboid is \(5\text{ cm} \times 6\text{ cm} \times 8\text{ cm} = 240\text{ cm}^3\). When the metal cuboid is completely submerged, the volume of water displaced is equal to the volume of the cuboid. The area of the container's base is \(15\text{ cm} \times 10\text{ cm} = 150\text{ cm}^2\). The rise in water level is the volume of displaced water divided by the base area of the container: \(240\text{ cm}^3 / 150\text{ cm}^2 = 1.6\text{ cm}\).

M1: for calculating the volume of the metal cuboid as \(5 \times 6 \times 8 = 240\). M1: for dividing their volume by the base area of the container (\(15 \times 10 = 150\)). A1: for 1.6.

First, express all terms as powers of 3: \(9 = 3^2\), \(27 = 3^3\), and \(81 = 3^4\). The equation becomes \(\frac{3^{n} \times (3^{2})^{n-1}}{(3^{3})^{2}} = 3^{4}\). Simplify the numerator and denominator: \(\frac{3^{n} \times 3^{2n-2}}{3^{6}} = 3^{4}\). Combine the powers of 3 on the left side: \(\frac{3^{3n-2}}{3^{6}} = 3^{4}\), which simplifies to \(3^{3n-8} = 3^{4}\). Since the bases are equal, equate the exponents: \(3n - 8 = 4\). Solve for \(n\): \(3n = 12\), which gives \(n = 4\).

M1: for expressing 9, 27, and 81 as powers of 3 (e.g. \(3^2\), \(3^3\), \(3^4\)). M1: for applying the laws of indices to obtain a single power of 3 on the left hand side, e.g. \(3^{3n-8} = 3^4\). A1: for \(n = 4\).

Since the probability of choosing a red counter is 0.4, the probability of choosing either a blue or a green counter is \(1 - 0.4 = 0.6\). Let the number of blue counters be \(2x\) and the number of green counters be \(3x\). The difference between the number of green and blue counters is \(3x - 2x = x\). We are told there are 9 more green counters than blue counters, so \(x = 9\). This means there are \(2 \times 9 = 18\) blue counters and \(3 \times 9 = 27\) green counters. The total number of blue and green counters combined is \(18 + 27 = 45\). Since blue and green counters represent 0.6 of the total number of counters \(T\), we have \(0.6 \times T = 45\). Solving for \(T\) gives \(T = 45 / 0.6 = 75\).

M1: for finding the difference in terms of a variable, e.g., \(3x - 2x = 9\), so \(x = 9\), and finding the total number of blue and green counters as 45. M1: for setting up the equation \(0.6 \times T = 45\) or equivalent ratio method. A1: for 75.

Since \(y\) is inversely proportional to the square of \(x\), we can write the equation as \(y = k / x^2\), where \(k\) is a constant. Substitute \(x = 3\) and \(y = 8\) to find \(k\): \(8 = k / 3^2\), which gives \(8 = k / 9\), so \(k = 72\). The equation is \(y = 72 / x^2\). Now substitute \(y = 18\) into the equation: \(18 = 72 / x^2\). Rearranging gives \(x^2 = 72 / 18 = 4\). Since \(x > 0\), we take the positive square root to get \(x = 2\).

M1: for setting up the proportional relationship \(y = k / x^2\) and substituting the initial values to find \(k = 72\). M1: for substituting \(y = 18\) into their equation to get \(x^2 = 4\). A1: for \(x = 2\).

The total number of matches is \(3 + 5 + x + 4 + 2 = 14 + x\). The total number of goals scored is \((0 \times 3) + (1 \times 5) + (2 \times x) + (3 \times 4) + (4 \times 2) = 0 + 5 + 2x + 12 + 8 = 25 + 2x\). Since the mean is 1.9, we have the equation \((25 + 2x) / (14 + x) = 1.9\). Multiply both sides by \((14 + x)\): \(25 + 2x = 1.9 \times (14 + x)\), which gives \(25 + 2x = 26.6 + 1.9x\). Subtract \(1.9x\) from both sides: \(25 + 0.1x = 26.6\). Subtract 25 from both sides: \(0.1x = 1.6\). Multiply by 10: \(x = 16\).

M1: for expressing the total number of matches as \(14 + x\) and the total number of goals as \(25 + 2x\). M1: for setting up the equation \((25 + 2x)/(14 + x) = 1.9\) and attempting to solve for \(x\). A1: for \(x = 16\).

Substitute the expression for \(y\) from the second equation into the first equation: \(2x^{2} - (2x - 5)^{2} = 17\). Expand the bracket: \(2x^{2} - (4x^{2} - 20x + 25) = 17\). Simplify the equation: \(2x^{2} - 4x^{2} + 20x - 25 = 17\), which gives \(-2x^{2} + 20x - 25 = 17\). Rearrange into a standard quadratic form: \(-2x^{2} + 20x - 42 = 0\). Divide the entire equation by -2: \(x^{2} - 10x + 21 = 0\). Factorise the quadratic equation: \((x - 3)(x - 7) = 0\). This gives two possible values for \(x\): \(x = 3\) or \(x = 7\). For \(x = 3\), substitute back to find \(y\): \(y = 2(3) - 5 = 1\). For \(x = 7\), substitute back to find \(y\): \(y = 2(7) - 5 = 9\). So the solutions are \(x = 3, y = 1\) or \(x = 7, y = 9\).

M1: for substituting the linear equation into the quadratic equation to get \(2x^2 - (2x-5)^2 = 17\). M1: for expanding and simplifying to get a quadratic equation in \(x\), e.g., \(x^2 - 10x + 21 = 0\). A1: for both pairs of correct values \(x = 3, y = 1\) and \(x = 7, y = 9\).

The area of a sector is given by \(\frac{\theta}{360} \times \pi r^{2}\). Given \(\theta = 60\) and Area = \(6\pi\), we have \(\frac{60}{360} \times \pi r^{2} = 6\pi\). This simplifies to \(\frac{1}{6} r^{2} = 6\), which gives \(r^{2} = 36\), so the radius \(r = 6\) cm. The perimeter of the sector is the sum of the two radii and the arc length: Perimeter = \(2r + \text{arc length}\). The arc length is \(\frac{60}{360} \times 2\pi r = \frac{1}{6} \times 12\pi = 2\pi\) cm. Therefore, the perimeter is \(2(6) + 2\pi = 12 + 2\pi\) cm.

M1: for setting up the equation for the area of the sector \(\frac{1}{6} \pi r^2 = 6\pi\) and finding \(r = 6\). M1: for calculating the arc length as \(\frac{1}{6} \times 2\pi \times 6 = 2\pi\). A1: for \(12 + 2\pi\).

First, find the total number of parts in the ratio: 2 + 3 + 5 = 10 parts. Next, calculate the volume of each ingredient needed for 4 litres: Orange juice is \(2/10 \times 4 = 0.8\) litres, Mango juice is \(3/10 \times 4 = 1.2\) litres, and Water is \(5/10 \times 4 = 2.0\) litres. Now work out the cost of each ingredient: Orange juice costs \(0.8 \times £1.20 = £0.96\); Mango juice costs \(1.2 \times £1.80 = £2.16\); Water is free. Total cost = \(£0.96 + £2.16 = £3.12\).

M1 for finding the volume of orange juice (0.8 litres) or mango juice (1.2 litres). M1 for multiplying their volume of orange juice by 1.20 or their volume of mango juice by 1.80. A1 for the correct total cost of £3.12 (accept 3.12).

To find when both buses leave at the same time again, we need the Lowest Common Multiple (LCM) of 12 and 18. Multiples of 12 are 12, 24, 36, 48, ... and multiples of 18 are 18, 36, 54, ... The LCM is 36 minutes. Adding 36 minutes to 9:00 am gives 9:36 am.

M1 for listing multiples of 12 and 18 to find a common multiple or using prime factorisation. M1 for finding the LCM is 36 minutes. A1 for 09:36 or 9:36 am.

The perimeter is given by \(2((2x + 5) + (x - 2)) = 42\). Simplifying inside the brackets gives \(2(3x + 3) = 42\), which expands to \(6x + 6 = 42\). Solving for \(x\) gives \(6x = 36\), so \(x = 6\). Thus, the length is \(2(6) + 5 = 17\) cm and the width is \(6 - 2 = 4\) cm. The area is \(\text{length} \times \text{width} = 17 \times 4 = 68\) cm\(^2\).

M1 for setting up a correct equation for perimeter, e.g. 2(2x + 5 + x - 2) = 42. M1 for solving to find x = 6 and substituting to find the dimensions. A1 for 68.

The volume of water in the cuboid is \(12 \times 5 \times 4 = 240\) cm\(^3\). The area of the right-angled triangular cross-section of the prism is \(0.5 \times 8 \times 6 = 24\) cm\(^2\). Since volume of a prism = area of cross-section \(\times\) length, we have \(24 \times L = 240\). Solving for \(L\) gives \(L = 10\) cm.

M1 for calculating the volume of the cuboid as 240. M1 for calculating the cross-sectional area as 24 and setting up the equation 24 * L = 240. A1 for 10.

The combined probability of picking a blue or green counter is \(1 - 0.35 = 0.65\). The ratio of blue to green is 2 : 3, so there are 5 parts in total. The probability of picking a blue counter is \(\frac{2}{5} \times 0.65 = 2 \times 0.13 = 0.26\).

M1 for finding the combined probability of blue and green, 1 - 0.35 = 0.65. M1 for dividing the combined probability by 5 and multiplying by 2. A1 for 0.26.

First, find the frequency \(x\) by subtracting the other frequencies from 25: \(x = 25 - (6 + 8 + 3 + 2) = 6\). Next, calculate the total goals scored: \((0 \times 6) + (1 \times 8) + (2 \times 6) + (3 \times 3) + (4 \times 2) = 0 + 8 + 12 + 9 + 8 = 37\). The mean number of goals is \(37 / 25 = 1.48\).

M1 for finding the missing frequency x = 6. M1 for calculating the total number of goals as 37. A1 for 1.48.

We can write \(y = \frac{k}{x^2}\). Substituting \(x = 3\) and \(y = 8\) gives \(8 = \frac{k}{3^2}\), so \(8 = \frac{k}{9}\), which means \(k = 72\). The equation is \(y = \frac{72}{x^2}\). When \(x = 6\), \(y = \frac{72}{6^2} = \frac{72}{36} = 2\).

M1 for setting up the relation y = k / x^2 and finding k = 72. M1 for substituting x = 6 into their formula. A1 for 2.

Substitute \(y = 2x - 5\) into the first equation: \(x^2 + (2x - 5)^2 = 25\). Expanding gives \(x^2 + 4x^2 - 20x + 25 = 25\). Simplifying gives \(5x^2 - 20x = 0\). Factorising gives \(5x(x - 4) = 0\). Thus \(x = 0\) or \(x = 4\). When \(x = 0\), \(y = 2(0) - 5 = -5\). When \(x = 4\), \(y = 2(4) - 5 = 3\). The solutions are \(x = 0, y = -5\) and \(x = 4, y = 3\).

M1 for substituting 2x - 5 to get a quadratic equation in x. M1 for factorising their simplified quadratic equation. A1 for both correct pairs of values.

Since 7.4 is rounded to 1 decimal place, the degree of accuracy is 0.1. The half-unit is \(0.1 / 2 = 0.05\). Lower bound: \(7.4 - 0.05 = 7.35\). Upper bound: \(7.4 + 0.05 = 7.45\). Thus, the error interval is \(7.35 \le y < 7.45\).

M1 for identifying 7.35 or 7.45 as a limit. A1 for the correct interval \(7.35 \le y < 7.45\) (accept alternative notation like \([7.35, 7.45)\)).

The multiplier for a 12% depreciation is \(1 - 0.12 = 0.88\). The value of the laptop after 2 years is \(650 \times 0.88^2 = 650 \times 0.7744 = 503.36\).

M1 for a complete method to calculate the depreciated value over 2 years, e.g., \(650 \times 0.88^2\) or calculating 12% first to get 572, then calculating 12% of 572. A1 for 503.36 (accept £503.36).

Expand the brackets: \(8x - 12 = 18\). Add 12 to both sides: \(8x = 30\). Divide both sides by 8: \(x = 30 / 8 = 3.75\).

M1 for a correct first algebraic step, e.g., expanding brackets to get \(8x - 12 = 18\) or dividing both sides by 4 to get \(2x - 3 = 4.5\). A1 for 3.75 (or equivalent fraction, e.g., \(\frac{15}{4}\)).

First find the radius of the semicircle: \(r = 12 / 2 = 6\text{ cm}\). The area of a semicircle is half the area of a circle: \(\text{Area} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \pi \times 6^2 = 18\pi \approx 56.5486...\text{ cm}^2\). To 3 significant figures, this is 56.5.

M1 for finding the radius is 6 and correctly identifying the formula for the area of a semicircle: \(\frac{1}{2} \times \pi \times 6^2\). A1 for 56.5 (accept answers in the range 56.5 to 56.6).

Expected frequency is calculated by multiplying the number of trials by the probability of the outcome: \(400 \times 0.35 = 140\).

M1 for \(400 \times 0.35\). A1 for 140.

First, calculate the total number of goals scored: \((0 \times 2) + (1 \times 4) + (2 \times 3) + (3 \times 1) = 0 + 4 + 6 + 3 = 13\). The total number of matches is \(2 + 4 + 3 + 1 = 10\). Mean = \(13 / 10 = 1.3\).

M1 for calculating the sum of products: \((0 \times 2) + (1 \times 4) + (2 \times 3) + (3 \times 1)\) or showing a sum of 13. A1 for 1.3.

Find the unit price for each brand. For Brand A: \(4.20 / 200 = 0.021\) per gram, or \(£2.10\) per 100g. For Brand B: \(5.10 / 250 = 0.0204\) per gram, or \(£2.04\) per 100g. Since \(£2.04 < £2.10\), Brand B is better value.

M1 for finding a comparable rate of cost per unit mass for at least one brand, e.g., \(4.20 / 2\) or \(5.10 / 2.5\). A1 for concluding Brand B with correct supporting calculations (such as £2.10 vs £2.04 per 100g).

The sequence increases by 7 each time, so the formula starts with \(7n\). The first term (where \(n=1\)) is 4, so we need to adjust by subtracting 3: \(7(1) - 3 = 4\). Therefore, the expression for the \(n\)-th term is \(7n - 3\).

M1 for writing \(7n + k\) where \(k\) is any integer (including \(k = 0\)), or showing a constant difference of 7. A1 for \(7n - 3\).

A number rounded to 1 decimal place has a maximum error of 0.05. The lower limit is 7.4 - 0.05 = 7.35 . The upper limit is 7.4 + 0.05 = 7.45 . Since the value is rounded up from 7.35 but must be strictly less than 7.45, the error interval is 7.35 <= x < 7.45 .

M1 for identifying the bounds 7.35 and 7.45. A1 for the correct interval 7.35 <= x < 7.45 (accept [7.35, 7.45) ).

First, find the cost of 1 pen: 3.80 / 5 = 0.76 pounds. Next, find the cost of 12 pens: 0.76 * 12 = 9.12 pounds.

M1 for a correct method to find the unitary cost 3.80 / 5 or to set up a proportion 3.80 * (12 / 5) . A1 for 9.12 (or 9.12 pounds).

Expand both brackets: 3(2x - 5) = 6x - 15 and 4(x + 2) = 4x + 8 . Collect like terms: (6x + 4x) + (-15 + 8) = 10x - 7 .

M1 for a correct expansion of at least one bracket (e.g., 6x - 15 or 4x + 8 ). A1 for 10x - 7 (or equivalent simplified form).

The diameter is 12 cm, so the radius r = 6 cm. The area of a circle is pi * r^2 , so the area of a semicircle is 0.5 * pi * r^2 . Area = 0.5 * pi * 6^2 = 18 * pi approx 56.54867... cm^2. Correct to 3 significant figures, this is 56.5.

M1 for a complete correct method to find the area of the semicircle using r = 6 (e.g., 0.5 * pi * 6^2 ). A1 for 56.5 (accept 56.5 to 56.6).

First, calculate the total number of goals: (0 * 2) + (1 * 4) + (2 * 3) + (3 * 1) = 0 + 4 + 6 + 3 = 13 goals. Then, divide by the total number of matches (10): 13 / 10 = 1.3 .

M1 for showing a correct method to find the total goals, e.g., (0 * 2) + (1 * 4) + (2 * 3) + (3 * 1) (or 13). A1 for 1.3.

The expected number of Heads is found by multiplying the total number of trials by the probability of landing on Heads: 200 * 0.65 = 130 .

M1 for 200 * 0.65 . A1 for 130.

The total amount after 2 years is 4000 * 1.025^2 = 4202.50 pounds. The interest earned is the final amount minus the principal investment: 4202.50 - 4000 = 202.50 pounds.

M1 for a correct method to find the total value or the total interest after 2 years, e.g., 4000 * 1.025^2 (or 4202.50). A1 for 202.50 (or 202.5).

The ratio of Dark : Milk : White is \(2 : 5 : 3\). The difference in parts between milk and white is \(5 - 3 = 2\) parts. We are given that 2 parts represent 24 chocolates. Therefore, 1 part represents \(24 / 2 = 12\) chocolates. The total number of parts is \(2 + 5 + 3 = 10\) parts. The total number of chocolates is \(10 \times 12 = 120\).

M1 for finding the difference in ratio parts \(5 - 3 = 2\) or setting up an equation such as \(5x - 3x = 24\). M1 for finding the value of one part (12) or the total parts (10). A1 for 120.

The perimeter of a rectangle is \(2 \times (\text{length} + \text{width})\). So, \(2 \times ((2x + 5) + (x + 3)) = 52\). Simplifying inside the brackets gives \(2 \times (3x + 8) = 52\). Expanding this gives \(6x + 16 = 52\). Subtracting 16 from both sides gives \(6x = 36\), which simplifies to \(x = 6\). Now we find the dimensions: Length = \(2(6) + 5 = 17\) cm, and Width = \(6 + 3 = 9\) cm. The area of the rectangle is \(17 \times 9 = 153\) cm\(^2\).

M1 for forming an equation for the perimeter, e.g., \(2(2x + 5 + x + 3) = 52\). M1 for solving the equation to find \(x = 6\). A1 for 153.

The perimeter consists of three sides of the rectangle (15 m, 8 m, and 15 m) and the curved boundary of the semi-circle. The straight edges sum to \(15 + 8 + 15 = 38\) m. The curved edge of the semi-circle is half the circumference of a circle of diameter 8 m: \(0.5 \times \pi \times 8 = 4\pi \approx 12.57\) m. Total perimeter = \(38 + 12.57 = 50.57\) m. To 1 decimal place, this is 50.6 m.

M1 for calculating the straight edge perimeter \(15 + 15 + 8 = 38\). M1 for calculating the curved arc length \(0.5 \times \pi \times 8\) (approx 12.57). A1 for 50.6.

A 15% reduction means the sale price is 85% of the original price. Let the original price be P. \(0.85 \times P = 74.80\). Solving for P gives \(P = 74.80 / 0.85 = 88\). The original price was £88.

M1 for recognizing 74.80 represents 85% or writing \(0.85 \times P = 74.80\). M1 for a complete method to find the original price, e.g., \(74.80 / 0.85\). A1 for 88.

The total probability of all outcomes is 1. The probability of picking either blue or yellow is \(1 - 0.3 = 0.7\). The ratio of blue to yellow is \(3 : 4\), which means there are \(3 + 4 = 7\) parts in total. The probability of picking a blue counter is \(\frac{3}{7}\) of the remaining probability: \(\frac{3}{7} \times 0.7 = 0.3\).

M1 for finding the remaining probability of blue or yellow (\(1 - 0.3 = 0.7\)). M1 for sharing 0.7 in the ratio \(3 : 4\), i.e., \(\frac{3}{7} \times 0.7\). A1 for 0.3.

Total number of matches = \(4 + 7 + x + 5 + 1 = 17 + x\). Total number of goals scored = \((0 \times 4) + (1 \times 7) + (2 \times x) + (3 \times 5) + (4 \times 1) = 0 + 7 + 2x + 15 + 4 = 2x + 26\). Since \(\text{Mean} = \frac{\text{Total Goals}}{\text{Total Matches}}\), we have: \(\frac{2x + 26}{17 + x} = 1.6\). Multiply both sides by \(17 + x\): \(2x + 26 = 1.6 \times (17 + x) \implies 2x + 26 = 27.2 + 1.6x\). Subtract 1.6x from both sides: \(0.4x + 26 = 27.2 \implies 0.4x = 1.2 \implies x = 3\).

M1 for expressing the total number of matches as \(17 + x\) or total goals as \(2x + 26\). M1 for setting up the equation \(\frac{2x + 26}{17 + x} = 1.6\). A1 for 3.

Value after the first year = \(25000 \times (1 - 0.12) = 25000 \times 0.88 = £22,000\). Let the multiplier for the second year's decrease be m. So, \(22000 \times m = 18700 \implies m = 18700 / 22000 = 0.85\). Since the multiplier is 0.85, the percentage decrease in the second year is \((1 - 0.85) \times 100 = 15\)%. Thus, x = 15.

M1 for calculating the value of the car after the first year as £22,000. M1 for dividing 18,700 by 22,000 to find the multiplier 0.85 or equivalent percentage remaining. A1 for 15.

Since the line is a tangent to the curve, equating the two equations gives \(x^2 + 5x + 7 = 3x + c\). Rearranging to form a quadratic equation: \(x^2 + 2x + (7 - c) = 0\). Since the line is a tangent, there is only one point of intersection, so the discriminant of this quadratic equation must be equal to 0 (\(b^2 - 4ac = 0\)). Here, \(a = 1\), \(b = 2\), and the constant term is \(7 - c\). Substituting these values: \(2^2 - 4 \times 1 \times (7 - c) = 0 \implies 4 - 4(7 - c) = 0 \implies 4 - 28 + 4c = 0 \implies -24 + 4c = 0 \implies 4c = 24 \implies c = 6\).

M1 for equating the equations to get \(x^2 + 2x + 7 - c = 0\). M1 for setting the discriminant \(b^2 - 4ac = 0\). A1 for 6.

First, calculate the volume of the water in the tank: \(\text{Volume} = 80\text{ cm} \times 60\text{ cm} \times 45\text{ cm} = 216,000\text{ cm}^3\). Next, convert the volume from cubic centimetres to litres: \(\text{Volume in litres} = \frac{216,000}{1000} = 216\text{ litres}\). Now, calculate the time taken to empty the tank at a rate of \(15\text{ litres}\) per minute: \(\text{Time} = \frac{216}{15} = 14.4\text{ minutes}\).

M1 for finding the volume of the water: \(80 \times 60 \times 45 = 216,000\). M1 for converting the volume to litres: \(\frac{216,000}{1000} = 216\) (or equivalent method). A1 for \(14.4\) (or equivalent, e.g., \(14\text{ minutes } 24\text{ seconds}\)).

First, write an equation for the perimeter of the rectangle: \(\text{Perimeter} = 2(3x + 4) + 2(2x - 1) = 56\). Expand the brackets and simplify: \(6x + 8 + 4x - 2 = 56\) which simplifies to \(10x + 6 = 56\). Solve for \(x\): \(10x = 50\) so \(x = 5\). Find the length and width of the rectangle: \(\text{Length} = 3(5) + 4 = 19\text{ cm}\) and \(\text{Width} = 2(5) - 1 = 9\text{ cm}\). Calculate the area of the rectangle: \(\text{Area} = 19 \times 9 = 171\text{ cm}^2\).

M1 for forming a correct equation for the perimeter, e.g., \(2(3x + 4) + 2(2x - 1) = 56\) or \(10x + 6 = 56\). M1 for solving the equation to find \(x = 5\). M1 for substituting \(x = 5\) to find the dimensions: \(19\) and \(9\). A1 for \(171\).

First, calculate the normal selling price to make a \(35\%\) profit: \(\text{Normal price} = 320 \times 1.35 = \pounds 432\). Next, apply the \(15\%\) reduction during the sale: \(\text{Sale price} = 432 \times (1 - 0.15) = 432 \times 0.85 = \pounds 367.20\).

M1 for finding the selling price before the discount: \(320 \times 1.35 = 432\). M1 for finding \(15\%\) of the selling price or multiplying by \(0.85\): \(432 \times 0.85\). A1 for \(367.20\) (accept \(367.2\)).

First, calculate the area of the triangle \(ABC\): \(\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 10 = 50\text{ cm}^2\). Next, use Pythagoras' theorem to find the length of the hypotenuse \(AC\): \(AC^2 = AB^2 + BC^2 = 10^2 + 10^2 = 200\), so \(AC = \sqrt{200}\text{ cm}\). The side \(AC\) is the diameter of the semicircle. Therefore, the radius \(r\) is \(r = \frac{\sqrt{200}}{2}\text{ cm}\). Calculate the area of the semicircle: \(\text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \left(\frac{\sqrt{200}}{2}\right)^2 = 25\pi \approx 78.54\text{ cm}^2\). Calculate the total area: \(\text{Total area} = 50 + 25\pi \approx 128.54\text{ cm}^2\). Rounding to 3 significant figures gives \(129\text{ cm}^2\).

M1 for finding the area of the triangle: \(\frac{1}{2} \times 10 \times 10 = 50\). M1 for using Pythagoras to find \(AC^2 = 200\) or \(AC \approx 14.14\). M1 for calculating the area of the semicircle: \(\frac{1}{2} \times \pi \times \left(\frac{\sqrt{200}}{2}\right)^2 \approx 78.54\). A1 for \(129\) (accept answers in the range \(128.5\) to \(129\)).

First, find the number of blue counters using the ratio \(R : B = 3 : 4\): Since there are 21 red counters, \(\text{Blue counters} = 21 \times \frac{4}{3} = 28\). The total number of red and blue counters is: \(21 + 28 = 49\). The probability of choosing a yellow counter is \(0.3\). This means the probability of choosing either a red or a blue counter is \(1 - 0.3 = 0.7\). Let \(T\) be the total number of counters in the bag. Therefore: \(0.7 \times T = 49\) which gives \(T = \frac{49}{0.7} = 70\).

M1 for finding the number of blue counters: \(21 \times \frac{4}{3} = 28\). M1 for finding the probability of red or blue: \(1 - 0.3 = 0.7\). M1 for setting up the equation \(0.7 T = 49\) or equivalent. A1 for \(70\).

To find the volume of the special offer bottle, calculate 18% of 750 ml and add it to the original volume: \(18\% \text{ of } 750 = 0.18 \times 750 = 135\) ml. Total volume = \(750 + 135 = 885\) ml. Alternatively, using a multiplier: \(750 \times 1.18 = 885\) ml.

M1 for a complete method to find the increased volume, e.g., \(750 \times 1.18\) or \(750 \times 0.18\) or 135 seen. A1 for 885.

First, expand the brackets on the left-hand side: \(5x - 15 = 28\). Next, add 15 to both sides of the equation: \(5x = 43\). Finally, divide both sides by 5: \(x = \frac{43}{5} = 8.6\).

M1 for correctly expanding the bracket to get \(5x - 15 = 28\) or for dividing both sides by 5 to get \(x - 3 = 5.6\). A1 for 8.6 or equivalent fraction.

First, convert £340 to Euros: \(340 \times 1.16 = 394.40\) Euros. Subtract the spent amount: \(394.40 - 250 = 144.40\) Euros. Then, convert the remaining Euros back to Pounds: \(144.40 \div 1.15 \approx 125.5652...\) Pounds. Rounding to the nearest penny gives £125.57.

M1 for finding the remaining Euros, \(340 \times 1.16 - 250\) (or 144.40), or for dividing their remaining Euros by 1.15. A1 for 125.57 (accept 125.56 to 125.57).

The perimeter of a semi-circle is made up of the curved arc and the straight diameter. Length of the curved arc = \(\frac{1}{2} \times \pi \times d = \frac{1}{2} \times \pi \times 12 = 6\pi \approx 18.85\) cm. Total perimeter = \(18.85 + 12 = 30.85\) cm. Rounded to 3 significant figures, this is 30.8 cm.

M1 for finding the arc length, e.g., \(\frac{1}{2} \times \pi \times 12\) (or 18.8...) or for \(\pi \times 6 + 12\). A1 for 30.8 (accept 30.8 - 30.9).

The total sum of all five numbers is \(16 \times 5 = 80\). The sum of the four given numbers is \(9 + 12 + 17 + 23 = 61\). The fifth number is \(80 - 61 = 19\).

M1 for a calculation to find the total sum, e.g., \(16 \times 5\) (or 80) or the sum of the four given numbers (or 61). A1 for 19.

The probability of the spinner landing on Green is \(1 - (0.45 + 0.25) = 1 - 0.70 = 0.3\). The estimated number of times the spinner lands on Green is \(150 \times 0.3 = 45\).

M1 for finding the probability of landing on Green, \(1 - (0.45 + 0.25)\) (or 0.3), or for finding the expected counts of Red and Blue (67.5 and 37.5). A1 for 45.

Divide the decimals: \(\frac{4.5}{1.25} = 3.6\). Subtract the powers of 10: \(10^7 \div 10^{-3} = 10^{7 - (-3)} = 10^{10}\). Combining these gives \(3.6 \times 10^{10}\).

M1 for \(3.6 \times 10^k\) where \(k \neq 10\), or showing \(36,000,000,000\), or demonstrating correct division of the numbers. A1 for \(3.6 \times 10^{10}\).

The number of years between the start of 2021 and the start of 2024 is 3 years. The multiplier for 12% depreciation is \(1 - 0.12 = 0.88\). Value at the start of 2024 = \(15000 \times 0.88^3 = 15000 \times 0.681472 = 10222.08\). To the nearest pound, the value is £10,222.

M1 for a complete method using the correct depreciation multiplier over 3 years, e.g., \(15000 \times 0.88^3\) or \(15000 \times 0.88^2\) (or 11616). A1 for 10222 (accept 10222.08).

When a number \( x \) is rounded to 1 decimal place, the degree of accuracy is 0.1. The maximum error is half of this, which is \( 0.05 \). Lower bound: \( 7.4 - 0.05 = 7.35 \). Upper bound: \( 7.4 + 0.05 = 7.45 \). Therefore, the error interval is \( 7.35 \le x < 7.45 \).

M1 for finding either the lower bound \( 7.35 \) or the upper bound \( 7.45 \). A1 for \( 7.35 \le x < 7.45 \).

Add 3 to both sides: \( 5y > 21 \). Divide both sides by 5: \( y > 4.2 \).

M1 for \( 5y > 21 \) or \( 5y = 21 \), or for an answer of \( 4.2 \) with an incorrect inequality sign or equals sign. A1 for \( y > 4.2 \) (or \( y > \frac{21}{5} \)).

First, multiply the map distance by the scale factor: \( 6.4 \times 25\,000 = 160\,000\text{ cm} \). Next, convert the distance from centimetres to metres: \( 160\,000 \div 100 = 1\,600\text{ m} \). Finally, convert metres to kilometres: \( 1\,600 \div 1000 = 1.6\text{ km} \).

M1 for \( 6.4 \times 25\,000 \) or \( 160\,000\text{ cm} \) or \( 1.6 \) with incorrect units. A1 for \( 1.6\text{ km} \) (or \( 1.6 \)).

First, find the radius \( r \) of the semicircle: \( r = \frac{12}{2} = 6\text{ cm} \). The area of a full circle is \( \pi r^2 \), so the area of a semicircle is half of that: \( \text{Area} = \frac{\pi \times 6^2}{2} = \frac{36\pi}{2} = 18\pi \). Using a calculator: \( 18\pi \approx 56.54866776\text{ cm}^2 \). Rounding to 3 significant figures gives \( 56.5\text{ cm}^2 \).

M1 for \( \pi \times 6^2 \div 2 \) or \( 18\pi \) or a decimal in the range \( 56.5 \) to \( 56.6 \). A1 for \( 56.5 \).

To find the expected number of times the coin lands on Heads, multiply the total number of spins by the probability of landing on Heads: \( 200 \times 0.65 = 130 \).

M1 for \( 200 \times 0.65 \). A1 for 130.

If the mean of five numbers is 12, their sum must be: \( 5 \times 12 = 60 \). Now, find the sum of the four given numbers: \( 8 + 15 + 11 + 14 = 48 \). Subtract the sum of the four numbers from the total sum to find the fifth number: \( 60 - 48 = 12 \).

M1 for \( 5 \times 12 \) or \( 60 \), or \( 8 + 15 + 11 + 14 = 48 \). A1 for 12.

The formula for compound interest is: \( A = P(1 + r)^t \), where \( P = 3000 \), \( r = 0.025 \), and \( t = 3 \). \( A = 3000 \times (1.025)^3 \). \( A = 3000 \times 1.076890625 \). \( A \approx 3230.671875 \). Rounding to the nearest penny gives \( £3230.67 \).

M1 for \( 3000 \times (1.025)^3 \) or \( 3230.67... \) or for showing year-by-year calculation steps with at least one correct year's interest added. A1 for 3230.67 (accept £3230.67).

Using the relationship for joint proportionality:

\(P = \frac{k x^2}{A}\), where \(k\) is a constant.

Substitute the given values to find \(k\):

\(25000 = \frac{k \times 500^2}{12000}\)

\(25000 = \frac{250000 k}{12000}\)

\(25000 = \frac{25 k}{1.2}\)

\(k = \frac{25000 \times 1.2}{25} = 1200\)

Now substitute \(k = 1200\), \(x = 800\), and \(A = 16000\) to find \(P\):

\(P = \frac{1200 \times 800^2}{16000}\)

\(P = \frac{1200 \times 640000}{16000}\)

\(P = 1200 \times 40 = 48000\)

Therefore, the profit is £48,000.

M1: Set up the correct proportionality equation \(P = \frac{k x^2}{A}\) or equivalent.
M1: Substitute \(x = 500\), \(A = 12000\), and \(P = 25000\) to find \(k = 1200\).
M1: Substitute \(k = 1200\), \(x = 800\), and \(A = 16000\) into the formula.
A1: Correct final answer of 48000 (or £48,000).

First, find the length of the diagonal of the square base, \(AC\).

Using Pythagoras' theorem:
\(AC = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \approx 14.142\text{ cm}\)

Since \(O\) is the centre of the base, the distance from corner \(A\) to \(O\) is half of the diagonal \(AC\):
\(AO = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \approx 7.071\text{ cm}\)

Triangle \(VOA\) is right-angled at \(O\). Using trigonometry:
\(\tan(62^\circ) = \frac{VO}{AO}\)

\(VO = AO \times \tan(62^\circ)\)

\(VO = 5\sqrt{2} \times \tan(62^\circ) \approx 7.071 \times 1.8807 \approx 13.2987\text{ cm}\)

Correct to 3 significant figures, the height \(VO\) is \(13.3\text{ cm}\).

M1: Use Pythagoras to find the diagonal \(AC = \sqrt{200}\) or half-diagonal \(AO = \sqrt{50} \approx 7.071\).
M1: Use a correct trigonometric ratio in triangle \(VOA\), e.g., \(\tan(62^\circ) = \frac{VO}{AO}\).
M1: Rearrange to find the height: \(VO = 5\sqrt{2} \tan(62^\circ)\).
A1: Correct answer of 13.3 (accept 13.3 to 13.30).

Multiply both sides by the common denominator \((2x - 3)(x + 5)\):

\(4(x + 5) + 3(2x - 3) = 2(2x - 3)(x + 5)\)

Expand the brackets on both sides:

\(4x + 20 + 6x - 9 = 2(2x^2 + 10x - 3x - 15)\)

\(10x + 11 = 2(2x^2 + 7x - 15)\)

\(10x + 11 = 4x^2 + 14x - 30\)

Rearrange into standard quadratic form \(ax^2 + bx + c = 0\):

\(4x^2 + 4x - 41 = 0\)

Use the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(x = \frac{-4 \pm \sqrt{4^2 - 4(4)(-41)}}{2(4)}\)

\(x = \frac{-4 \pm \sqrt{16 + 656}}{8}\)

\(x = \frac{-4 \pm \sqrt{672}}{8}\)

Using the positive root:
\(x = \frac{-4 + 25.92296}{8} \approx 2.74\)

Using the negative root:
\(x = \frac{-4 - 25.92296}{8} \approx -3.74\)

The solutions are \(x = 2.74\) or \(x = -3.74\).

M1: Eliminate the denominators by multiplying through, yielding \(4(x+5) + 3(2x-3) = 2(2x-3)(x+5)\) or equivalent.
M1: Expand and simplify to obtain a 3-term quadratic equation, e.g., \(4x^2 + 4x - 41 = 0\).
M1: Apply the quadratic formula correctly to their quadratic equation.
A1: Correct solutions \(x = 2.74\) and \(x = -3.74\) (accept either order, allow "2.74, -3.74").

The formula for compound interest is:

\(A = P\left(1 + \frac{r}{100}\right)^t\)

Substituting the given values:

\(8790.02 = 8000\left(1 + \frac{r}{100}\right)^3\)

Divide by 8000:

\(\left(1 + \frac{r}{100}\right)^3 = \frac{8790.02}{8000} \approx 1.0987525\)

Take the cube root of both sides:

\(1 + \frac{r}{100} = \sqrt[3]{1.0987525} \approx 1.032\)

Subtract 1:

\(\frac{r}{100} = 0.032\)

\(r = 3.2\)

The interest rate is \(3.2\%\).

M1: Set up the compound interest equation, e.g., \(8000(1 + \frac{r}{100})^3 = 8790.02\) or \(8000x^3 = 8790.02\).
M1: Rearrange to find \((1 + \frac{r}{100})^3\) or \(x^3 \approx 1.0987525\).
M1: Solve for \(r\) by taking the cube root, e.g., \(1 + \frac{r}{100} \approx 1.032\).
A1: Correct answer of 3.2.

Let the total number of marbles in the bag be \(n\).

The number of blue marbles is \(0.45n\). Since the number of blue marbles must be an integer, \(n\) must be a multiple of 20 (since \(0.45 = \frac{9}{20}\)).

The number of red marbles is 5.

Therefore, the number of green marbles is:
\(n - 0.45n - 5 = 0.55n - 5\)

The probability of choosing two green marbles without replacement is:
\(P(\text{Green, Green}) = \frac{0.55n - 5}{n} \times \frac{0.55n - 6}{n - 1} = \frac{3}{38}\)

Let's test \(n = 20\) (the smallest possible integer value for \(n\)):

Number of green marbles = \(0.55(20) - 5 = 11 - 5 = 6\).

\(P(\text{Green, Green}) = \frac{6}{20} \times \frac{5}{19} = \frac{30}{380} = \frac{3}{38}\).

This perfectly matches the given probability! Hence, \(n = 20\).

Now, calculate the number of blue marbles:
\(\text{Blue} = 0.45 \times 20 = 9\).

M1: Express the number of green marbles in terms of \(n\), i.e., \(0.55n - 5\).
M1: Write down the correct probability equation for drawing two green marbles without replacement: \(\frac{0.55n-5}{n} \times \frac{0.55n-6}{n-1} = \frac{3}{38}\).
M1: Solve the equation or systematically test valid multiples of 20 to find \(n = 20\).
A1: Correctly calculate the number of blue marbles as 9.

For the interval \(10 < t \le 15\):
- Class Width = \(15 - 10 = 5\)
- Frequency = 20
- Frequency Density (FD) = \(\frac{20}{5} = 4\)

We are given that this bar has a height of 8 cm. This means the scale on the frequency density axis is:
\(\text{Height} = 2 \times \text{FD}\) (or \(1\text{ unit of FD} = 2\text{ cm}\)).

For the interval \(15 < t \le 25\):
- Class Width = \(25 - 15 = 10\)
- Height = 5 cm
- Frequency Density (FD) = \(\frac{\text{Height}}{2} = \frac{5}{2} = 2.5\)

Using the formula \(\text{Frequency} = \text{FD} \times \text{Class Width}\):
\(x = 2.5 \times 10 = 25\).

M1: Calculate the frequency density for the interval \(10 < t \le 15\) as 4.
M1: Establish the relationship between height and frequency density (Height = \(2 \times \text{FD}\) or equivalent).
M1: Use this relationship to find the FD for the interval \(15 < t \le 25\) as 2.5.
A1: Correctly calculate \(x = 25\).

Let the number of part-time workers be \(3k\) and the number of full-time workers be \(5k\).

First, calculate the daily earnings for one individual of each type:
- Daily pay for a part-time worker: \(4\text{ hours} \times £12/\text{hour} = £48\)
- Daily pay for a full-time worker: \(8\text{ hours} \times £15/\text{hour} = £120\)

Now, express the total wage bill in terms of \(k\):
\(\text{Total Wages} = 3k(48) + 5k(120)\)
\(\text{Total Wages} = 144k + 600k = 744k\)

We are given that the total wage bill is £2232:
\(744k = 2232\)
\(k = \frac{2232}{744} = 3\)

The total number of workers is:
\(3k + 5k = 8k = 8 \times 3 = 24\).

M1: Calculate the daily pay of a part-time worker (£48) or a full-time worker (£120).
M1: Set up an expression or equation for the total wage bill, e.g., \(3k \times 48 + 5k \times 120 = 2232\).
M1: Solve the equation to find \(k = 3\) or find the number of part-time workers (9) and full-time workers (15).
A1: Correctly calculate the total number of workers as 24.

First, calculate the area of the rectangle:
\(\text{Area}_{\text{rectangle}} = 12 \times 8 = 96\text{ m}^2\)

Next, calculate the area of the semicircle. Since the diameter is 8 m, the radius \(r = 4\text{ m}\):
\(\text{Area}_{\text{semicircle}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (4^2) = 8\pi \approx 25.1327\text{ m}^2\)

Now find the total area of the garden:
\(\text{Total Area} = 96 + 25.1327 = 121.1327\text{ m}^2\)

Calculate the total cost of the patio tiles:
\(\text{Total Cost} = 121.1327 \times £18.50 \approx £2240.95\)

To the nearest pound, the total cost is £2241.

M1: Calculate the area of the rectangle as 96.
M1: Calculate the area of the semicircle as \(8\pi\) or approximately 25.13.
M1: Sum the areas and multiply by £18.50, e.g., \((96 + 8\pi) \times 18.5\).
A1: Correctly round the final answer to 2241 (accept £2241).

First, convert the Paris price into pounds: €75.90 / 1.15 = £66.00. Next, convert the New York price into pounds: $85.80 / 1.32 = £65.00. Comparing the prices in pounds: London is £68.00, Paris is £66.00, and New York is £65.00. The cheapest price is £65.00 (New York) and the most expensive price is £68.00 (London). The difference is £68.00 - £65.00 = £3.00.

M1 for converting Paris price to pounds: 75.90 / 1.15 (= 66) or New York price to pounds: 85.80 / 1.32 (= 65). M1 for converting both prices correctly to pounds (£66 and £65). M1 for identifying the cheapest (£65) and most expensive (£68) prices. A1 for 3 or £3 or £3.00.

Area of the rectangle = \((2x + 3)(x - 1) = 2x^2 + x - 3\). Area of the square = \((x + 1)^2 = x^2 + 2x + 1\). We are given that Area of rectangle - Area of square = 16. So, \((2x^2 + x - 3) - (x^2 + 2x + 1) = 16\). Simplifying this gives: \(x^2 - x - 4 = 16\) which rearranges to \(x^2 - x - 20 = 0\). Factorising the quadratic equation: \((x - 5)(x + 4) = 0\). Since length and width must be positive, \(x\) must be greater than 1, so we choose the positive solution \(x = 5\).

M1 for expressing the area of the rectangle as \((2x + 3)(x - 1)\) and expanding to \(2x^2 + x - 3\) (allow one algebraic slip). M1 for setting up the correct equation \((2x^2 + x - 3) - (x^2 + 2x + 1) = 16\) or equivalent. M1 for reducing to \(x^2 - x - 20 = 0\) and attempting to solve by factorisation. A1 for \(x = 5\) (rejecting \(x = -4\) as a final answer).

The volume of the cylinder is \(V = \pi r^2 h = \pi \times 3^2 \times 10 = 90\pi \text{ cm}^3\). Since \(10\%\) of the metal is lost, \(90\%\) remains: \(0.9 \times 90\pi = 81\pi \text{ cm}^3\). The volume of one sphere is \(V_{sphere} = \frac{4}{3}\pi R^3 = \frac{4}{3} \times \pi \times 1.5^3 = 4.5\pi \text{ cm}^3\). The number of spheres that can be made is \(81\pi / 4.5\pi = 18\).

M1 for finding the volume of the cylinder: \(\pi \times 3^2 \times 10\) or \(90\pi\) (or approximately 282.7). M1 for finding the remaining volume after 10% loss: \(0.90 \times 90\pi\) or \(81\pi\) (or approximately 254.5). M1 for finding the volume of one sphere: \(\frac{4}{3} \times \pi \times 1.5^3\) or \(4.5\pi\) (or approximately 14.1). A1 for 18.

For the first interval \(0 < t \le 10\): Class width = 10, Frequency = 15. Therefore, Frequency Density = \(15 / 10 = 1.5\). Since the height of this bar is \(6 \text{ cm}\), the scale for the frequency density axis is \(6 / 1.5 = 4 \text{ cm}\) of height per unit of frequency density. For the second interval \(10 < t \le 25\): Class width = 15. The height of the bar is \(8 \text{ cm}\). Using the scale, the frequency density represented by \(8 \text{ cm}\) height is \(8 / 4 = 2\). Therefore, the Frequency = Frequency Density \(\times\) Class Width = \(2 \times 15 = 30\).

M1 for finding the frequency density of the first interval: \(15 / 10 = 1.5\). M1 for finding the relationship between height and frequency density: e.g., \(6 \text{ cm} / 1.5 = 4 \text{ cm}\) per unit of frequency density. M1 for calculating the frequency density of the second interval as \(8 / 4 = 2\) and multiplying by its class width of 15. A1 for 30.

The probability of choosing a lemon sweet first is \(\frac{6}{n}\). Since there is no replacement, the probability of choosing a second lemon sweet is \(\frac{5}{n-1}\). The probability that both are lemon is \(\frac{6}{n} \times \frac{5}{n-1} = \frac{30}{n(n-1)}\). Equating this to the given probability: \(\frac{30}{n(n-1)} = \frac{1}{3}\). Multiplying both sides by \(3n(n-1)\) gives \(90 = n(n-1)\). This expands to \(n^2 - n - 90 = 0\). Factorising the quadratic equation gives \((n - 10)(n + 9) = 0\). Since the total number of sweets must be a positive integer, \(n = 10\).

M1 for writing an expression for the probability of selecting two lemon sweets: \(\frac{6}{n} \times \frac{5}{n-1}\). M1 for setting up the equation \(\frac{30}{n(n-1)} = \frac{1}{3}\) and cross-multiplying to obtain \(n(n-1) = 90\) or \(n^2 - n - 90 = 0\). M1 for solving the quadratic equation by factorisation: \((n-10)(n+9)=0\). A1 for \(n = 10\) (must reject \(n = -9\)).