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First, find the total cost of the 15 packs of pens: \(15 \times £2.40 = £36.00\). Next, subtract this total cost from the \(£50\) note to find the change: \(£50 - £36 = £14\).

M1 for a correct method to find the total cost, e.g. \(15 \times 2.40\) (\(36\)) M1 for a correct subtraction method, e.g. \(50 - \text{their } 36\) A1 for \(£14\) (accept \(14\))

First, find the fraction of students studying French and German: \(\frac{2}{7} + \frac{1}{2} = \frac{4}{14} + \frac{7}{14} = \frac{11}{14}\). The fraction of students studying Spanish is the remainder: \(1 - \frac{11}{14} = \frac{3}{14}\). Since this represents 30 students, we can set up the equation: \(\frac{3}{14} \text{ of total} = 30\). Therefore, the total number of students is \(30 \div \frac{3}{14} = 30 \times \frac{14}{3} = 140\).

M1 for a correct method to add the two fractions, e.g. finding a common denominator of 14 to get \(\frac{11}{14}\) M1 for subtracting their fraction from 1 to find the fraction for Spanish, e.g. \(1 - \frac{11}{14} = \frac{3}{14}\), and equating to 30 A1 for 140

The difference in the ratio parts between Fred and Dan is \(9 - 2 = 7\) parts. This difference corresponds to 56 stamps. So, 1 part is equal to \(56 \div 7 = 8\) stamps. The total number of ratio parts is \(2 + 5 + 9 = 16\) parts. Therefore, the total number of stamps shared is \(16 \times 8 = 128\).

M1 for finding the difference in ratio parts, e.g. \(9 - 2 = 7\) parts, and setting it equal to 56 M1 for a complete method to find the total number of stamps, e.g. \((2 + 5 + 9) \times (56 \div 7)\) A1 for 128

The perimeter of the triangle is the sum of all three sides: \(2(2x - 3) + (x + 4) = 38\). Expand the brackets: \(4x - 6 + x + 4 = 38\). Simplify the equation: \(5x - 2 = 38\). Add 2 to both sides: \(5x = 40\). Divide by 5 to find \(x\): \(x = 8\).

M1 for setting up a correct expression for the perimeter, e.g. \(2(2x - 3) + (x + 4) = 38\) M1 for simplifying the equation to the form \(ax = b\), e.g. \(5x = 40\) A1 for 8

First, find the height of the cuboid. The area of the square base is \(4 \times 4 = 16\text{ cm}^2\). Volume = Base Area \(\times\) Height, so \(16 \times h = 120\), which gives \(h = \frac{120}{16} = 7.5\text{ cm}\). The cuboid has 2 square faces of area \(16\text{ cm}^2\) and 4 rectangular side faces of area \(4 \times 7.5 = 30\text{ cm}^2\). The total surface area is \(2 \times 16 + 4 \times 30 = 32 + 120 = 152\text{ cm}^2\).

M1 for finding the height of the cuboid, e.g. \(120 \div 16 = 7.5\) M1 for a correct method to calculate the total surface area using their height, e.g. \(2 \times (4 \times 4) + 4 \times (4 \times 7.5)\) A1 for 152

First, calculate the total height of the girls: \(4 \times 1.50 = 6.00\text{ m}\). Next, calculate the total height of the boys: \(6 \times 1.65 = 9.90\text{ m}\). The overall total height for all 10 children is \(6.00 + 9.90 = 15.90\text{ m}\). Finally, find the mean height of the group: \(15.90 \div 10 = 1.59\text{ m}\).

M1 for finding the total height of either the girls (\(6.00\)) or the boys (\(9.90\)) M1 for dividing the combined total height by 10, e.g. \((6.00 + 9.90) \div 10\) A1 for 1.59

The probability that the coin lands on Tails is \(1 - 0.35 = 0.65\). The estimated number of times the coin lands on Tails is \(0.65 \times 200 = 130\). Alternatively, the expected number of Heads is \(0.35 \times 200 = 70\). The expected number of Tails is \(200 - 70 = 130\).

M1 for finding the probability of Tails, e.g. \(1 - 0.35 = 0.65\), OR finding the expected number of Heads, e.g. \(0.35 \times 200 = 70\) M1 for completing the calculation, e.g. \(0.65 \times 200\) or \(200 - 70\) A1 for 130

First, expand the first bracket: \(3 \times 2x - 3 \times 1 = 6x - 3\). Next, expand the second bracket (being careful with the signs): \(-2 \times x - 2 \times (-4) = -2x + 8\). Combine the expanded terms: \(6x - 3 - 2x + 8\). Simplify by collecting like terms: \(6x - 2x = 4x\) and \(-3 + 8 = 5\). This gives the final simplified expression: \(4x + 5\).

M1 for expanding the first bracket correctly to get \(6x - 3\) M1 for expanding the second bracket correctly to get \(-2x + 8\) (or showing a subtraction of \(2x - 8\)) A1 for \(4x + 5\)

First, calculate the multiplication in the numerator:
\(1.5 \times 7 = 10.5\)

Next, add the terms in the numerator:
\(4.5 + 10.5 = 15\)

Finally, divide the result by the denominator:
\(\frac{15}{0.3} = \frac{150}{3} = 50\)

M1: For evaluating \(1.5 \times 7 = 10.5\)
M1: For adding 4.5 to their 10.5 to get 15 and attempting to divide by 0.3
A1: 50

Find the number of adults:
\(\frac{3}{8} \text{ of } 240 = (240 \div 8) \times 3 = 30 \times 3 = 90\)

Find the number of teenagers:
\(10\% \text{ of } 240 = 24\)
\(40\% \text{ of } 240 = 24 \times 4 = 96\)
\(5\% \text{ of } 240 = 12\)
\(45\% \text{ of } 240 = 96 + 12 = 108\)

Find the total number of adults and teenagers:
\(90 + 108 = 198\)

Find the number of children:
\(240 - 198 = 42\)

M1: For finding the number of adults (90) OR finding the number of teenagers (108)
M1: For adding their number of adults and teenagers (e.g. \(90 + 108 = 198\)) and subtracting from 240
A1: 42

Find the difference in parts between green and red marbles:
\(7 - 3 = 4 \text{ parts}\)

Since this difference represents 24 marbles:
\(1 \text{ part} = 24 \div 4 = 6 \text{ marbles}\)

Find the total number of parts in the ratio:
\(3 + 5 + 7 = 15 \text{ parts}\)

Multiply the total parts by the value of one part:
\(15 \times 6 = 90\)

M1: For finding the difference in ratio parts between green and red, e.g. \(7 - 3 = 4\) parts
M1: For dividing 24 by their difference in parts to find the value of one part (e.g. \(24 \div 4 = 6\))
A1: 90

Expand the bracket on the left side:
\(8x - 12 = 2x + 18\)

Subtract \(2x\) from both sides:
\(6x - 12 = 18\)

Add 12 to both sides:
\(6x = 30\)

Divide by 6:
\(x = 5\)

M1: For expanding the bracket correctly to obtain \(8x - 12\)
M1: For collecting terms in \(x\) on one side and numerical constants on the other, e.g. \(8x - 2x = 18 + 12\)
A1: 5 (or \(x = 5\))

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

So, the highest common factor is \(6xy\).

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

\(\frac{18xy^2}{6xy} = 3y\)

Write the fully factorised expression:
\(6xy(2x - 3y)\)

M1: For identifying any common factor (e.g. \(2\), \(3\), \(x\), \(y\), \(xy\), \(3x\)) from the terms, e.g. \(3(4x^2y - 6xy^2)\) or \(xy(12x - 18y)\)
M1: For pulling out a common factor of 6 and at least one variable, e.g. \(6x(2xy - 3y^2)\) or \(6y(2x^2 - 3xy)\)
A1: \(6xy(2x - 3y)\)

First, find the probability of landing on either a 5 or a 6:
\(P(5 \text{ or } 6) = 0.3 + 0.15 = 0.45\)

Next, multiply the total number of rolls by this probability to find the expected frequency:
\(\text{Estimate} = 200 \times 0.45\)

\(200 \times 0.45 = 2 \times 45 = 90\)

M1: For adding the probabilities: \(0.3 + 0.15 = 0.45\)
M1: For multiplying their probability by 200, e.g. \(200 \times 0.45\)
A1: 90

Calculate the total number of goals by multiplying each number of goals by its frequency:
- \(0 \times 4 = 0\)
- \(1 \times 7 = 7\)
- \(2 \times 5 = 10\)
- \(3 \times 3 = 9\)
- \(4 \times 1 = 4\)

Sum of goals:
\(0 + 7 + 10 + 9 + 4 = 30\)

Total number of matches (sum of frequencies):
\(4 + 7 + 5 + 3 + 1 = 20\)

Calculate the mean:
\(\text{Mean} = \frac{30}{20} = 1.5\)

M1: For attempting to calculate at least three correct products of goals and frequency (e.g., finding at least three of 0, 7, 10, 9, 4)
M1: (dep) For dividing their total sum of goals (e.g. 30) by 20
A1: 1.5

First, calculate the area of the triangular cross-section:
\(\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}\)
\(\text{Area of triangle} = \frac{1}{2} \times 6 \times 8 = 24\text{ cm}^2\)

Next, calculate the volume of the prism by multiplying the cross-sectional area by the length of the prism:
\(\text{Volume} = 24 \times 15\)
\(24 \times 15 = 360\text{ cm}^3\)

M1: For a correct method to find the area of the triangular face, e.g. \(\frac{1}{2} \times 6 \times 8\) (or 24)
M1: For multiplying their cross-sectional area by 15 (e.g. \(24 \times 15\))
A1: 360

First, find the total number of parts in the ratio: \(2 + 3 + 5 = 10\). Next, find the value of one part: \(\pounds 140 \div 10 = \pounds 14\). Alice's initial share is \(2 \times \pounds 14 = \pounds 28\). Charlie's initial share is \(5 \times \pounds 14 = \pounds 70\). Alice gives half of her share to Charlie, which is \(\pounds 28 \div 2 = \pounds 14\). Charlie's new total is \(\pounds 70 + \pounds 14 = \pounds 84\).

M1 for finding the value of one share: \(140 \div 10 = 14\) or showing Alice's share is \(\pounds 28\) and Charlie's share is \(\pounds 70\). M1 for finding half of Alice's share (\(\pounds 14\)) or attempting to add half of Alice's share to Charlie's. A1 for 84 (accept \pounds 84).

Evaluate the numerator first: \(5^2 \times 3 - 15 = 25 \times 3 - 15 = 75 - 15 = 60\). Next, evaluate the denominator: \(2^3 + 4 = 8 + 4 = 12\). Finally, divide the numerator by the denominator: \(60 \div 12 = 5\).

M1 for evaluating \(5^2 \times 3 - 15 = 75 - 15\) or obtaining 60 for the numerator. M1 for evaluating \(2^3 + 4 = 8 + 4\) or obtaining 12 for the denominator. A1 for 5.

First, calculate the number of students who play football: \(\frac{3}{8} \times 240 = 3 \times 30 = 90\). Next, calculate the number of students who play tennis: 35% of 240. Since 10% of 240 = 24, then 30% is 72, and 5% is 12. So, 35% of 240 = 72 + 12 = 84. The number of students playing rugby is \(240 - (90 + 84) = 240 - 174 = 66\).

M1 for finding the number of football players: 90. M1 for finding the number of tennis players: 84. A1 for 66.

First expand the first bracket: \(4 \times 2x - 4 \times 3 = 8x - 12\). Then expand the second bracket, taking care with the negative sign: \(-3 \times x - 3 \times (-5) = -3x + 15\). Combine the terms: \(8x - 12 - 3x + 15 = (8x - 3x) + (-12 + 15) = 5x + 3\).

M1 for expanding the first bracket correctly: \(8x - 12\). M1 for expanding the second bracket correctly: \(-3x + 15\) (allow one sign error, e.g. \(-3x - 15\) for method mark if the process is clear). A1 for \(5x + 3\).

First, expand the bracket on the left side of the equation: \(6x - 3 = 4x + 9\). Next, subtract \(4x\) from both sides: \(2x - 3 = 9\). Then, add 3 to both sides: \(2x = 12\). Finally, divide both sides by 2: \(x = 6\).

M1 for expanding the bracket correctly: \(6x - 3\). M1 for isolating the x terms on one side and the number terms on the other side, e.g. \(6x - 4x = 9 + 3\) or \(2x = 12\) (follow through their expansion). A1 for 6 (or x = 6).

Since BCD is a straight line, the angles ACB and ACD lie on a straight line and sum to 180 degrees. Therefore, angle \(ACB = 180^\circ - 130^\circ = 50^\circ\). The angles in triangle ABC sum to 180 degrees. So, \(4x + (2x + 10) + 50 = 180\). Simplify the equation: \(6x + 60 = 180\). Subtract 60 from both sides: \(6x = 120\). Divide by 6: \(x = 20\).

M1 for finding angle \(ACB = 180^\circ - 130^\circ = 50^\circ\) or writing the equation \(4x + 2x + 10 = 130\). M1 for setting up the equation \(4x + 2x + 10 + 50 = 180\) or \(6x + 60 = 180\) (or \(6x + 10 = 130\)). A1 for 20 (or x = 20).

First, find the total score for the girls: \(4 \times 15 = 60\). Next, find the total score for the boys: \(6 \times 10 = 60\). Find the combined total score for all 10 children: \(60 + 60 = 120\). Finally, divide this total by the total number of children to find the overall mean: \(120 \div 10 = 12\).

M1 for finding the total score of the girls (60) or the boys (60). M1 for finding the combined total score (120) and dividing by the total number of children (10). A1 for 12.

First, find the probability that the coin lands on Tails: \(1 - 0.65 = 0.35\). To find the expected number of times it lands on Tails, multiply this probability by the number of spins: \(300 \times 0.35\). Since \(300 \times 0.35 = 3 \times 35 = 105\), the expected number of Tails is 105.

M1 for finding the probability of Tails: \(1 - 0.65 = 0.35\) or finding the expected number of Heads: \(300 \times 0.65 = 195\). M1 for a complete method to find expected Tails: \(300 \times 0.35\) or \(300 - 195\). A1 for 105.

The ratio is Alice : Bob : Charlie = \(3 : 5 : 4\). The difference in ratio parts between Bob and Alice is \(5 - 3 = 2\) parts. Since Bob receives \(24\) more than Alice, these \(2\) parts represent \(24\). Therefore, \(1\) part is worth \(24 \div 2 = 12\). The total number of parts is \(3 + 5 + 4 = 12\) parts. The total amount of money shared is \(12 \times 12 = 144\).

M1 for finding the difference in ratio parts between Bob and Alice: \(5 - 3 = 2\) parts (or setting up an equation such as \(5x - 3x = 24\)). M1 for finding the value of one part: \(24 \div 2 = 12\), or for summing the parts \(3 + 5 + 4 = 12\) and showing a method to find the total. A1 for \(144\) (or \(144\)).

The wall consists of a rectangle and a triangle. First, find the area of the rectangle: \(\text{Area} = 6\text{ m} \times 3\text{ m} = 18\text{ m}^2\). Next, find the height of the triangle: \(\text{Height} = 5\text{ m} - 3\text{ m} = 2\text{ m}\). The base of the triangle is the same as the length of the rectangle, which is \(6\text{ m}\). Find the area of the triangle: \(\text{Area} = \frac{1}{2} \times 6\text{ m} \times 2\text{ m} = 6\text{ m}^2\). Finally, add the two areas together for the total area: \(18\text{ m}^2 + 6\text{ m}^2 = 24\text{ m}^2\).

M1 for finding the area of the rectangle: \(6 \times 3 = 18\) OR for finding the height of the triangle: \(5 - 3 = 2\). M1 for a complete method to find the area of the triangle: \(\frac{1}{2} \times 6 \times \text{their height}\). A1 for \(24\) (ignore units).

To solve the equation \(\frac{5x - 2}{3} = 6\), first multiply both sides by \(3\): \(5x - 2 = 18\). Next, add \(2\) to both sides: \(5x = 20\). Finally, divide both sides by \(5\): \(x = 4\).

M1 for multiplying both sides by \(3\) to get \(5x - 2 = 18\). M1 for adding \(2\) to both sides of their equation (e.g., \(5x = 20\)). A1 for \(4\) (or \(x = 4\)).

First, calculate the total amount spent by the customer:
\(100 - 31.20 = 68.80\) (total spent in \(£\))

Next, calculate the cost of the 8 bags of compost:
\(8 \times 6.45 = 51.60\) (spent on compost in \(£\))

Subtract this from the total spent to find the cost of the plant pots:
\(68.80 - 51.60 = 17.20\) (spent on plant pots in \(£\))

Finally, divide by the price of one plant pot to find the quantity bought:
\(17.20 \div 2.15 = 8\)

M1: Method to find total spent, e.g. \(100 - 31.20\) (or \(68.80\) seen) or method to find cost of compost bags, e.g. \(8 \times 6.45\) (or \(51.60\) seen).
M1: Method to find total spent on plant pots, e.g. \(68.80 - 51.60\) (or \(17.20\) seen).
A1: 8 (accept 8 plant pots).

First, convert both capacities to the same unit so we can compare them:
Bottle A: \(750\text{ ml}\)
Bottle B: \(1.25\text{ litres} = 1250\text{ ml}\)

Now, calculate the cost per \(100\text{ ml}\) for each:

For Bottle A:
\(750\text{ ml} \div 100 = 7.5\) portions of \(100\text{ ml}\).
Cost per \(100\text{ ml} = 2.40 \div 7.5 = £0.32\) (or \(32\text{ p}\)).

For Bottle B:
\(1250\text{ ml} \div 100 = 12.5\) portions of \(100\text{ ml}\).
Cost per \(100\text{ ml} = 3.80 \div 12.5 = £0.304\) (or \(30.4\text{ p}\)).

Comparing the prices: \(30.4\text{ p} < 32\text{ p}\), so Bottle B is better value.

M1: Correct conversion of \(1.25\text{ l}\) to \(1250\text{ ml}\) or both costs written in pence.
M1: Attempt to find comparable values for both bottles, e.g. \(2.40 \div 7.5\) and \(3.80 \div 12.5\) (or equivalent division to find unit price).
A1: Correct figures of \(32\text{ p}\) (or \(£0.32\)) and \(30.4\text{ p}\) (or \(£0.304\)) with 'Bottle B' selected as the better value.

To calculate the mean, we first find the total number of goals scored by multiplying each number of goals by its frequency:

\(0 \times 3 = 0\)
\(1 \times 6 = 6\)
\(2 \times 5 = 10\)
\(3 \times 4 = 12\)
\(4 \times 2 = 8\)

Total number of goals = \(0 + 6 + 10 + 12 + 8 = 36\)
Total number of matches = \(3 + 6 + 5 + 4 + 2 = 20\)

Mean number of goals = \(\frac{36}{20} = 1.8\)

M1: At least 3 correct products of \(x \times f\) shown (can be implied by addition, e.g. 6, 10, 12, 8).
M1: (dep) Division of total goals (36) by total matches (20), i.e., \(36 \div 20\).
A1: 1.8

First expand the bracket on the left-hand side:
\(5x + 15 = 2x + 27\)

Next, subtract \(2x\) from both sides to collect the \(x\) terms on one side:
\(3x + 15 = 27\)

Subtract 15 from both sides to isolate the term with \(x\):
\(3x = 12\)

Finally, divide by 3:
\(x = 4\)

M1: Correct expansion of bracket, e.g. \(5x + 15\).
M1: Correct method to isolate the term in \(x\) on one side and constant on other side, e.g. \(5x - 2x = 27 - 15\) (allow error in one term during rearrangement).
A1: 4 (accept \(x = 4\)).

First, calculate the price of the laptop including VAT:
\(450 \times 1.20 = 540\) (or \(20\%\) of \(450 = 90\), so \(450 + 90 = 540\) in \(£\)).

Next, apply the \(15\%\) reduction to the price including VAT:
\(15\%\) of \(540 = 0.15 \times 540 = 81\)
Sale price = \(540 - 81 = 459\) (or \(540 \times 0.85 = 459\) in \(£\)).

M1: Correct method to find the price including VAT, e.g. \(450 \times 1.2\) or \(450 + 90\) (or \(540\) seen).
M1: Correct method to reduce their price including VAT by \(15\%\), e.g. \(\text{their } 540 \times 0.85\) or \(\text{their } 540 - (0.15 \times \text{their } 540)\).
A1: 459 (accept \(£459\)).

First, find the total area of the rectangle:
\(\text{Area of rectangle} = 12 \times 8 = 96\text{ cm}^2\)

Next, find the area of the semi-circle. Since the diameter is \(6\text{ cm}\), the radius \(r = 3\text{ cm}\):
\(\text{Area of semi-circle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \pi \times 3^2 = 4.5\pi \approx 14.137\text{ cm}^2\)

Subtract the area of the semi-circle from the area of the rectangle:
\(\text{Remaining area} = 96 - 14.137 = 81.863\text{ cm}^2\)

Rounding to 1 decimal place gives \(81.9\text{ cm}^2\).

M1: Correctly calculates area of rectangle as \(96\text{ cm}^2\).
M1: Correct method to calculate the area of the semi-circle of diameter \(6\text{ cm}\), e.g. \(\frac{1}{2} \times \pi \times 3^2\) (or \(14.1\) or \(14.14\) seen).
A1: 81.9 (accept answers in range \(81.86\) to \(81.9\)).

First, calculate the probability of landing on Yellow. The sum of all probabilities is 1:
\(P(\text{Yellow}) = 1 - (0.25 + 0.35 + 0.15)\)
\(P(\text{Yellow}) = 1 - 0.75 = 0.25\)

Next, calculate the expected frequency by multiplying the probability of Yellow by the number of spins:
\(\text{Expected value} = 240 \times 0.25 = 60\)

M1: Correct method to find the sum of known probabilities or subtract them from 1, e.g. \(1 - (0.25 + 0.35 + 0.15)\) (or \(0.25\) seen as probability of Yellow).
M1: (dep) Multiplies their probability of Yellow by 240, e.g. \(\text{their } 0.25 \times 240\).
A1: 60

First, find the gradient \(m\) of the line:
\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2\)

Now use the equation \(y = mx + c\) with one of the points, say \((2, 3)\), to find the y-intercept \(c\):
\(3 = 2(2) + c\)
\(3 = 4 + c\)
\(c = -1\)

Therefore, the equation of the line is:
\(y = 2x - 1\)

M1: Correct method to find the gradient, e.g. \(\frac{11 - 3}{6 - 2}\) (or gradient = 2 seen).
M1: Correct method to find \(c\) by substituting their gradient and coordinate of one of the points into \(y = mx + c\).
A1: \(y = 2x - 1\) (or equivalent equation, e.g., \(y - 3 = 2(x - 2)\)).

Number of packs needed from Shop A = 140 / 20 = 7 packs. Cost for Shop A = 7 * £4.50 = £31.50. Number of packs needed from Shop B = 140 / 35 = 4 packs. Cost for Shop B = 4 * £7.20 = £28.80. Shop B is cheaper by £31.50 - £28.80 = £2.70.

M1 for calculating the cost of 140 cards from one shop (e.g. 7 * 4.50 = 31.50 or 4 * 7.20 = 28.80) or finding the number of packs needed for both (7 and 4). M1 for calculating both costs correctly (£31.50 and £28.80). A1 for Shop B and £2.70.

Convert the total mass to grams: 4.8 kg = 4800 g. Total ratio parts = 13 + 7 = 20 parts. Mass of one part = 4800 / 20 = 240 g. Mass of copper = 13 * 240 = 3120 g.

M1 for converting 4.8 kg to 4800 g or working out mass of copper in kg (e.g. 4.8 / 20 * 13 = 3.12 kg). M1 for (4800 / 20) * 13 or (4.8 / 20) * 13. A1 for 3120.

The new price represents 115% of the original price. Original price = 241.50 / 1.15 = £210.

M1 for identifying that £241.50 corresponds to 115%. M1 for 241.50 / 1.15. A1 for 210.

The perimeter of a rectangle is 2 * (length + width). So, 2 * ((3x + 2) + (2x - 5)) = 54. Simplify inside brackets: 2 * (5x - 3) = 54. 10x - 6 = 54. 10x = 60. x = 6.

M1 for writing an expression for the perimeter, e.g., 2(3x + 2 + 2x - 5) or 3x + 2 + 3x + 2 + 2x - 5 + 2x - 5. M1 for simplifying and equating to 54, e.g., 10x - 6 = 54. A1 for x = 6.

The sum of probabilities is 1. Probability of blue or green = 1 - 0.35 = 0.65. The ratio of blue to green is 2 : 3, giving 2 + 3 = 5 parts. 1 part = 0.65 / 5 = 0.13. Probability of green = 3 * 0.13 = 0.39.

M1 for subtracting the red probability from 1: 1 - 0.35 = 0.65. M1 for dividing the remaining probability by the total parts: 0.65 / 5 (= 0.13) or 0.65 / 5 * 3. A1 for 0.39.

The perimeter of a semicircle consists of the curved arc plus the straight diameter. Radius r = 8 / 2 = 4 m. Curved arc length = \(\pi \times r = \pi \times 4 \approx 12.566\) m. Total perimeter = 12.566 + 8 = 20.57 m (to 2 decimal places).

M1 for calculating the curved arc length: \(\pi \times 4\) or \(\pi \times 8 / 2\) (approx 12.57). M1 for adding the diameter (8) to their arc length: \((\pi \times 4) + 8\). A1 for 20.57 (accept 20.56 to 20.57 from use of pi).

Calculate the total number of goals: (0 * 6) + (1 * 8) + (2 * 5) + (3 * 4) + (4 * 2) = 0 + 8 + 10 + 12 + 8 = 38 goals. Total matches = 6 + 8 + 5 + 4 + 2 = 25. Mean = 38 / 25 = 1.52 goals.

M1 for finding at least three correct products of goals * frequency (e.g. 8, 10, 12, 8). M1 for sum of products divided by 25: 38 / 25. A1 for 1.52.

The sum of the interior angles of a pentagon is (5 - 2) * 180 = 540 degrees. Sum of the given four angles = 115 + 130 + 95 + 120 = 460 degrees. Therefore, x = 540 - 460 = 80.

M1 for calculating the sum of the interior angles of a pentagon: (5 - 2) * 180 = 540. M1 for 540 - (115 + 130 + 95 + 120). A1 for 80.

First, convert the time into hours.
2 hours and 45 minutes is equivalent to
\(2 + \frac{45}{60} = 2.75\) hours.

Next, use the formula for average speed:
\(\text{Average Speed} = \frac{\text{Distance}}{\text{Time}}\)
\(\text{Average Speed} = \frac{143}{2.75} = 52\) mph.

M1: Convert 2 hours and 45 minutes to 2.75 hours (or write a correct method using minutes: \(\frac{143}{165} \times 60\)).
M1: Divide 143 by their decimal time (e.g., \(\frac{143}{2.75}\)).
A1: 52.

First, find the total number of siblings:
\((0 \times 5) + (1 \times 8) + (2 \times 5) + (3 \times 2) = 0 + 8 + 10 + 6 = 24\) siblings.

The total number of students is given as 20.

Now, calculate the mean:
\(\text{Mean} = \frac{24}{20} = 1.2\)

M1: For a correct method to find the total number of siblings, showing at least three products summed (e.g., \(0 + 8 + 10 + 6\) or \(24\)).
M1: (dep) For dividing their total number of siblings by 20.
A1: 1.2

Evaluate the numerator:
\(\sqrt{148.6 - 32.7} = \sqrt{115.9} \approx 10.765686...\)

Evaluate the denominator:
\(5.3 \times 1.4 = 7.42\)

Divide the numerator by the denominator:
\(\frac{10.765686...}{7.42} \approx 1.450901...\)

Rounding to 2 decimal places gives 1.45.

M1: Find the value of the numerator \(\approx 10.77\) or the denominator \(7.42\).
M1: Show an unrounded answer of at least 4 decimal places (e.g., \(1.4509...\)).
A1: 1.45

Expand the bracket on the left side:
\(5x - 15 = 2x + 9\)

Subtract \(2x\) from both sides:
\(3x - 15 = 9\)

Add 15 to both sides:
\(3x = 24\)

Divide by 3:
\(x = 8\)

M1: Expand the bracket correctly to get \(5x - 15\).
M1: Rearrange the equation to isolate terms in \(x\) on one side and numerical terms on the other (e.g., \(3x = 24\) or \(5x - 2x = 9 + 15\)).
A1: 8

First, find the area of the triangular cross-section:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 6 \times 4.5 = 13.5\text{ cm}^2\)

Next, calculate the volume by multiplying the cross-sectional area by the length of the prism:
\(\text{Volume} = 13.5 \times 15 = 202.5\text{ cm}^3\)

M1: For a correct method to find the cross-sectional area of the triangle: \(0.5 \times 6 \times 4.5\) (or \(13.5\)).
M1: (dep) For multiplying their cross-sectional area by 15.
A1: 202.5

Calculate the price after the first 20% reduction:
\(650 \times 0.80 = 520\) (or \(650 - 130 = 520\)).

Calculate the final price after the second 5% reduction:
\(520 \times 0.95 = 494\) (or \(520 - 26 = 494\)).

The final price of the television is £494.

M1: For calculating the first sale price: \(650 \times 0.8\) or 520.
M1: (dep) For applying a 5% reduction to their first sale price: \(\text{their } 520 \times 0.95\).
A1: 494

The sum of all probabilities is 1.
\(P(A) + P(B) + P(C) + P(D) = 1\)
\(0.4 + 0.3 + P(C) + P(D) = 1\)
\(P(C) + P(D) = 0.3\)

Since \(P(D)\) is twice \(P(C)\), let \(P(C) = x\) and \(P(D) = 2x\).
\(x + 2x = 0.3\)
\(3x = 0.3\)
\(x = 0.1\)

Therefore, \(P(D) = 2x = 2 \times 0.1 = 0.2\).

M1: For working out the combined probability of C and D: \(1 - (0.4 + 0.3) = 0.3\).
M1: For setting up an equation to find \(P(C)\) or \(P(D)\), e.g., \(x + 2x = 0.3\) or dividing 0.3 by 3.
A1: 0.2

Expand the first bracket:
\(3 \times 2x - 3 \times 5 = 6x - 15\)

Expand the second bracket, being careful with the signs:
\(-2 \times x - 2 \times (-4) = -2x + 8\)

Combine the expressions:
\(6x - 15 - 2x + 8\)

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

This gives:
\(4x - 7\)

M1: Expands the first bracket correctly to get \(6x - 15\).
M1: Expands the second bracket correctly to get \(-2x + 8\).
A1: 4x - 7

To find the scale factor for 20 muffins from a recipe of 8 muffins:

\(\text{Scale factor} = \frac{20}{8} = 2.5\)

Now, calculate the required amount of each ingredient for 20 muffins:
- Flour: \(250\text{ g} \times 2.5 = 625\text{ g}\)
- Sugar: \(120\text{ g} \times 2.5 = 300\text{ g}\)
- Eggs: \(2 \times 2.5 = 5\text{ eggs}\)

Comparing the required amounts to what Harry has:
- Harry has 600 g of flour, but needs 625 g (Not enough flour)
- Harry has 350 g of sugar, and needs 300 g (Enough sugar)
- Harry has 6 eggs, and needs 5 eggs (Enough eggs)

Since he does not have enough flour, he cannot make 20 muffins.

M1: For calculating the scaling factor of \(2.5\) (or equivalent, e.g., finding the weight of flour per muffin as \(31.25\text{ g}\)).
M1: For calculating the correct required amount for at least one ingredient for 20 muffins (e.g., \(625\text{ g}\) of flour, \(300\text{ g}\) of sugar, or \(5\) eggs).
A1: For 'No' with a correct supporting comparison showing that \(625\text{ g}\) of flour is needed but only \(600\text{ g}\) is available.

First, find the price after the 15% reduction:
\(\text{Reduction} = 0.15 \times 240 = £36\)
\(\text{Reduced price} = 240 - 36 = £204\)
(Alternatively: \(240 \times 0.85 = £204\))

Next, find the price after this reduced price is increased by 15%:
\(\text{Increase} = 0.15 \times 204 = £30.60\)
\(\text{Final price} = 204 + 30.60 = £234.60\)
(Alternatively: \(204 \times 1.15 = £234.60\))

M1: For a correct method to find the reduced price, e.g., \(240 \times 0.85\) or \(240 - 36\) (implied by \(£204\)).
M1: For a correct method to find the final price by increasing their reduced price by 15%, e.g., \(\text{their } 204 \times 1.15\) or \(\text{their } 204 + 30.60\).
A1: For \(£234.60\) (accept \(234.6\) or \(234.60\)).

First, find the radius of the cylinder:
\(\text{Radius} = \frac{\text{Diameter}}{2} = \frac{8}{2} = 4\text{ cm}\)

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

Substitute the values into the formula:
\(V = \pi \times 4^2 \times 15\)
\(V = \pi \times 16 \times 15\)
\(V = 240\pi \approx 753.982...\text{ cm}^3\)

Rounding to 3 significant figures gives \(754\text{ cm}^3\).

M1: For finding the radius is \(4\text{ cm}\) and substituting into the volume formula \(\pi \times r^2 \times 15\).
M1: For evaluating \(\pi \times 4^2 \times 15\) or \(240\pi\) or \(753.98...\)
A1: For \(754\) (accept answers in the range \(753.6\) to \(754.1\)).

First, find the probability \(x\) of landing on 4. The sum of all probabilities must equal 1:
\(0.15 + 0.35 + 0.2 + x = 1\)
\(0.7 + x = 1\)
\(x = 0.3\)

Next, calculate the expected number of times it lands on 4 in 300 spins:
\(\text{Estimate} = 300 \times 0.3 = 90\)

M1: For a correct method to find \(x\), e.g., \(1 - (0.15 + 0.35 + 0.2)\) or \(0.3\).
M1: For multiplying their \(x\) by 300, where \(0 < x < 1\), e.g., \(300 \times 0.3\).
A1: For \(90\).

First, find the percentage of students who are not in Year 7 or Year 8:
\(100\% - 45\% = 55\%\)

Now, calculate \(55\%\) of 760:
\(0.55 \times 760 = 418\)

Alternatively, calculate the number of students in Year 7 or Year 8:
\(0.45 \times 760 = 342\)
Subtract this from the total:
\(760 - 342 = 418\)

M1: For a correct method to find 55% of 760 (e.g. \(0.55 \times 760\)) OR finding 45% of 760 (342)
M1 (dep): For subtracting their 342 from 760
A0.5: 418

To convert from Pounds to Euros, multiply the amount in Pounds by the exchange rate:
\(350 \times 1.14 = 399\)

So, Liam receives €399.

M1.5: For a correct method to convert Pounds to Euros, i.e. \(350 \times 1.14\)
A1: 399

First, find the total number of goals scored by multiplying each number of goals by its frequency and summing them:
\((0 \times 3) + (1 \times 6) + (2 \times 7) + (3 \times 4) = 0 + 6 + 14 + 12 = 32\) goals

Next, divide the total number of goals by the total number of matches (the sum of frequencies):
\(\text{Total matches} = 3 + 6 + 7 + 4 = 20\)

\(\text{Mean} = \frac{32}{20} = 1.6\)

M1: For finding the total number of goals: \(0\times3 + 1\times6 + 2\times7 + 3\times4 = 32\)
M1: For dividing their total goals by 20
A0.5: 1.6

The perimeter of a semi-circle consists of the curved arc plus the straight diameter.

1. Calculate the length of the curved arc (which is half of the circumference of a circle with diameter 12 cm):
\(\text{Arc length} = \frac{\pi \times d}{2} = \frac{\pi \times 12}{2} = 6\pi \approx 18.85\text{ cm}\)

2. Add the diameter to get the total perimeter:
\(\text{Perimeter} = 18.85 + 12 = 30.85\text{ cm}\)

Rounding to 1 decimal place gives 30.8 cm.

M1: For a correct method to find the curved arc length: \(\pi \times 12 \div 2\) (or approx 18.8)
M1: For adding the diameter 12 to their curved arc length
A0.5: 30.8 (accept 30.85 or 30.84 with correct working if unrounded)

Multiply both sides of the equation by 2:
\(3x - 5 = 16\)

Add 5 to both sides:
\(3x = 21\)

Divide by 3:
\(x = 7\)

M1: For multiplying by 2 on both sides: \(3x - 5 = 16\)
M1: For isolating the x-term: \(3x = 21\)
A0.5: 7

First, expand both parts of the expression:
\(4(2x + 3) = 8x + 12\)
\(-3(x - 2) = -3x + 6\)

Combine the terms:
\(8x + 12 - 3x + 6\)

Collect like terms:
\((8x - 3x) + (12 + 6) = 5x + 18\)

M1: For expanding at least one bracket correctly to get either \(8x + 12\) or \(-3x + 6\) (note: watch for sign error on \(-3 \times -2\))
M1: For fully expanding to \(8x + 12 - 3x + 6\)
A0.5: \(5x + 18\) (or equivalent simplified form)

First, find the probability of choosing a green counter. Since the probabilities of all outcomes must add up to 1:
\(P(\text{green}) = 1 - (0.35 + 0.4) = 1 - 0.75 = 0.25\)

Now, multiply this probability by the total number of counters to find the number of green counters:
\(0.25 \times 60 = 15\)

Alternatively, find the number of red and blue counters first:
Number of red counters = \(0.35 \times 60 = 21\)
Number of blue counters = \(0.4 \times 60 = 24\)
Number of green counters = \(60 - (21 + 24) = 60 - 45 = 15\)

M1: For finding the probability of selecting a green counter: \(1 - (0.35 + 0.4) = 0.25\) OR for finding the total number of red and blue counters (45)
M1: For multiplying their probability of green by 60: \(0.25 \times 60\) OR for subtracting their red and blue total from 60: \(60 - 45\)
A0.5: 15

We want to find the first term where \(5n - 3 > 100\).

Let's set up an inequality:
\(5n - 3 > 100\)
\(5n > 103\)
\(n > 20.6\)

Since \(n\) must be an integer, the first term greater than 100 is the 21st term (\(n = 21\)).

Calculate the 21st term:
\(5(21) - 3 = 105 - 3 = 102\)

Alternatively, we can test values of \(n\) close to 20:
For \(n = 20\): \(5(20) - 3 = 97\)
For \(n = 21\): \(5(21) - 3 = 102\)
Thus, 102 is the first term greater than 100.

M1: For setting up the inequality \(5n - 3 > 100\) or trial and improvement around \(n = 20\)
M1: For evaluating both the 20th term (97) and 21st term (102)
A0.5: 102

First, convert the cost in Paris from euros to pounds by dividing by the exchange rate: \(78 \div 1.18 = 66.10169...\) pounds. This is \(£66.10\) to the nearest penny. Next, find the difference between this price and the London price: \(66.10169... - 64 = 2.10169...\) pounds. To the nearest penny, this is \(£2.10\).

M1 for \(78 \div 1.18\) or \(64 \times 1.18\) (finding Paris price in pounds or London price in euros). M1 for finding the difference between the two prices in the same currency (e.g., \(66.10 - 64\) or \(78 - 75.52\)). A1 for \(£2.10\) (accept 2.10).

First, calculate the numerator: \(18.5 - 2.9 = 15.6\), and \(\sqrt{15.6} \approx 3.9496835\). Next, calculate the denominator: \(0.4^3 = 0.064\). Now divide the numerator by the denominator: \(\frac{3.9496835}{0.064} \approx 61.7138...\) Rounding to 3 significant figures gives \(61.7\).

M1 for finding the value of the numerator as \(\approx 3.95\) or the denominator as \(0.064\). M1 for \(61.7138...\) shown as an intermediate step. A1 for \(61.7\).

Multiply each number of goals by its frequency to find the total goals: \(0 \times 5 = 0\), \(1 \times 8 = 8\), \(2 \times 4 = 8\), \(3 \times 3 = 9\). Sum these values to get total goals: \(0 + 8 + 8 + 9 = 25\). Divide the total goals by the total number of matches (20): \(\frac{25}{20} = 1.25\).

M1 for finding the sum of products: \(0 \times 5 + 1 \times 8 + 2 \times 4 + 3 \times 3\) (implied by 25). M1 for dividing their total goals by 20. A1 for \(1.25\).

Calculate the interest earned in one year: \(4500 \times 0.024 = 108\). Multiply by 3 to find the total interest over 3 years: \(108 \times 3 = 324\). Add the total interest to the original investment: \(4500 + 324 = 4824\).

M1 for \(4500 \times 0.024\) (or 108) or \(4500 \times 0.024 \times 3\) (or 324). M1 for adding their total interest to 4500. A1 for \(4824\) (accept 4824).

First, find the probability of landing on D. Since total probability is 1: \(P(D) = 1 - (0.15 + 0.42 + 0.18) = 1 - 0.75 = 0.25\). To find the expected frequency, multiply this probability by the number of spins: \(300 \times 0.25 = 75\).

M1 for \(1 - (0.15 + 0.42 + 0.18)\) (or 0.25). M1 for \(300 \times 0.25\) (or \(300 \times \text{their } P(D)\)). A1 for \(75\).

Expand the brackets: \(8x - 12 = 14\). Add 12 to both sides: \(8x = 26\). Divide by 8: \(x = \frac{26}{8} = 3.25\).

M1 for expanding brackets correctly to \(8x - 12 = 14\) or dividing both sides by 4 to get \(2x - 3 = 3.5\). M1 for isolating the x term (e.g., \(8x = 26\) or \(2x = 6.5\)). A1 for \(3.25\) (accept equivalent fractions such as \(\frac{13}{4}\)).

Substitute the values into the formula: \(v^2 = 5.8^2 + 2(3.5)(18)\). Calculate each term: \(5.8^2 = 33.64\) and \(2 \times 3.5 \times 18 = 126\). Add the terms: \(v^2 = 33.64 + 126 = 159.64\). Take the square root: \(v = \sqrt{159.64} \approx 12.63487...\) Rounding to 1 decimal place gives \(12.6\).

M1 for substituting the values correctly: \(5.8^2 + 2 \times 3.5 \times 18\). M1 for evaluating to \(159.64\) and attempting the square root. A1 for \(12.6\).

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

M1 for \(\pi \times 4.5^2 \times 12\). M1 for evaluating as \(243\pi\) or \(763.4...\) A1 for \(763\) (accept values in the range 763 to 764).

First, convert the mass of flour into grams: \(1.2\text{ kg} = 1200\text{ g}\). The ratio of flour to sugar is \(5 : 2\). Let's check which ingredient is the limiting factor: If we were to use all \(1200\text{ g}\) of flour, the sugar required would be \(1200 \times \frac{2}{5} = 480\text{ g}\). Since we only have \(450\text{ g}\) of sugar, sugar is the limiting factor. If we use all \(450\text{ g}\) of sugar, the flour required is \(450 \times \frac{5}{2} = 1125\text{ g}\). This is possible since we have \(1200\text{ g}\) of flour. The maximum total mass of the mixture is \(1125\text{ g} + 450\text{ g} = 1575\text{ g}\).

M1: Converting \(1.2\text{ kg}\) to \(1200\text{ g}\) or attempting to find the scaling factor (e.g., \(450 \div 2 = 225\) or \(1200 \div 5 = 240\)). M1: Complete calculation to find the mass of flour needed for \(450\text{ g}\) of sugar (\(1125\text{ g}\)) or calculating the total mixture parts (\(225 \times 7\)). A0.5: Correct answer of 1575.

The sale price of \(\text{£}357\) represents \(100\% - 15\% = 85\%\) of the original normal price. Let the normal price be \(x\). We can write this as: \(0.85 \times x = 357\). Solving for \(x\): \(x = \frac{357}{0.85} = 420\). Therefore, the normal price was \(\text{£}420\).

M1: Recognising that \(85\%\) corresponds to \(357\) (e.g., writing \(0.85x = 357\) or \(357 \div 85\)). M1: A complete method to calculate the original price (e.g., \(\frac{357}{0.85}\) or \(\frac{357}{85} \times 100\)). A0.5: Correct answer of 420.

First, calculate the volume of water needed to reach a depth of \(30\text{ cm}\): \(\text{Volume} = 80\text{ cm} \times 50\text{ cm} \times 30\text{ cm} = 120,000\text{ cm}^3\). Next, convert this volume to litres: \(\text{Volume in litres} = \frac{120,000}{1000} = 120\text{ litres}\). Finally, calculate the time required: \(\text{Time} = \frac{120\text{ litres}}{4\text{ litres per minute}} = 30\text{ minutes}\).

M1: Correctly calculating the required volume in cubic centimetres: \(80 \times 50 \times 30 = 120,000\text{ cm}^3\). M1: Converting their volume into litres and dividing by 4 to find the time (e.g., \(120,000 \div 1000 \div 4\)). A0.5: Correct answer of 30.

Let the smallest integer be \(n\). The next two consecutive integers are \(n + 1\) and \(n + 2\). Summing these together: \(n + (n + 1) + (n + 2) = 138\). Simplifying the equation gives: \(3n + 3 = 138\). Subtract 3 from both sides: \(3n = 135\). Divide by 3: \(n = 45\). The smallest of the three integers is 45.

M1: Setting up a correct algebraic equation for the consecutive integers, e.g., \(n + (n+1) + (n+2) = 138\), or an equivalent trial and improvement method showing three consecutive numbers summing near 138. M1: Correct simplification to \(3n = 135\) and solving for \(n\). A0.5: Correct answer of 45.

The sum of all probabilities is \(1\). Therefore, the probability of picking a yellow counter is: \(\text{P(yellow)} = 1 - (0.35 + 0.4) = 1 - 0.75 = 0.25\). Let the total number of counters be \(N\). We know that \(0.25 \times N = 10\). Solving for \(N\): \(N = \frac{10}{0.25} = 40\). There are 40 counters in total.

M1: Finding the probability of a yellow counter: \(1 - (0.35 + 0.4) = 0.25\). M1: Using the yellow probability to set up the calculation for total counters, e.g., \(10 \div 0.25\) or \(10 \times 4\). A0.5: Correct answer of 40.

First, expand each bracket separately. For the first bracket: \(4(2x - 3) = 8x - 12\). For the second bracket, including the negative sign: \(-3(x - 5) = -3x + 15\). Now, combine and simplify the like terms: \((8x - 3x) + (-12 + 15) = 5x + 3\).

M1: Correctly expanding at least one of the brackets (e.g., \(8x - 12\) or \(-3x + 15\)). M1: Showing the fully expanded expression with correct signs: \(8x - 12 - 3x + 15\). A0.5: Correct simplified expression of \(5x + 3\).

First, find the total number of books read by multiplying each book category by its frequency: \(1 \times 4 = 4\), \(2 \times 6 = 12\), \(3 \times 5 = 15\), \(4 \times 3 = 12\), \(5 \times 2 = 10\). Summing these together: \(\text{Total books} = 4 + 12 + 15 + 12 + 10 = 53\). The total number of classmates is \(20\). Therefore, the mean is: \(\text{Mean} = \frac{53}{20} = 2.65\).

M1: Correct method to calculate the total number of books (sum of products), e.g., \(4 + 12 + 15 + 12 + 10 = 53\). M1: Dividing their sum of products by the total frequency \(20\). A0.5: Correct answer of 2.65.

The sum of the interior angles of a pentagon is given by the formula \((n - 2) \times 180^\circ\). For \(n = 5\), the sum is: \((5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ\). Next, sum the four known interior angles: \(110^\circ + 125^\circ + 95^\circ + 130^\circ = 460^\circ\). Subtract this total from the sum of all interior angles to find the fifth angle: \(540^\circ - 460^\circ = 80^\circ\).

M1: Finding the sum of the interior angles of a pentagon to be \(540^\circ\) or using \((5 - 2) \times 180\). M1: Summing the four given interior angles to get \(460^\circ\). A0.5: Correct answer of 80.

Find the scale factor: 15 divided by 6 = 2.5. Multiply the original amount of butter by this scale factor: 150 grams multiplied by 2.5 = 375 grams. Alternatively, find the butter required for 1 pancake: 150 grams divided by 6 = 25 grams. Then multiply by 15: 25 grams multiplied by 15 = 375 grams.

M1: For a correct method to find the scaling factor (e.g. \(15 \div 6\)) or the amount of butter per pancake (e.g. \(150 \div 6\)). M1: For multiplying their intermediate value by the target number (dep on first M1). A0.5: Correct answer 375.

To find the expected frequency, multiply the total number of trials by the probability of the outcome: Expected value = \(140 \times 0.35 = 49\).

M1: For writing a correct expression to find the expected value, e.g. \(140 \times 0.35\) or equivalent fraction form. M1: Show step of multiplication or simplified expression. A0.5: Correct final answer of 49.

The volume of a prism is given by: Volume = area of cross-section multiplied by length. Volume = \(18.5\text{ cm}^2 \times 8.2\text{ cm} = 151.7\text{ cm}^3\).

M1: For a correct substitution into the prism volume formula: \(18.5 \times 8.2\). A1: For the correct numerical value 151.7. B0.5: For stating the correct units of volume, \(\text{cm}^3\).

Divide both sides by 4.2: \(x - 1.5 = 14.7 \div 4.2\) which gives \(x - 1.5 = 3.5\). Add 1.5 to both sides: \(x = 3.5 + 1.5 = 5\).

M1: For a correct first step of expanding brackets (e.g. \(4.2x - 6.3 = 14.7\)) or dividing by 4.2 (e.g. \(x - 1.5 = 3.5\)). M1: For isolating the x term (e.g. \(4.2x = 21\) or \(x = 3.5 + 1.5\)). A0.5: Correct final answer 5.

Find the reduction: \(15\%\) of \(420 = 0.15 \times 420 = 63\). Subtract the reduction from the original price: \(420 - 63 = 357\). Alternative method: Multiply the original price by the multiplier for a \(15\%\) reduction: \(420 \times 0.85 = 357\).

M1: For a correct method to find the reduction amount (e.g. \(420 \times 0.15\)) or a correct percentage multiplier (e.g. \(0.85\)). M1: For subtracting their correct reduction from 420 or completing the multiplier calculation. A0.5: Correct answer 357 (accept £357).

Step 1: Find the sum of (Goals Scored multiplied by Frequency): \((0 \times 5) + (1 \times 8) + (2 \times 4) + (3 \times 3) = 0 + 8 + 8 + 9 = 25\). Step 2: Find the total frequency: \(5 + 8 + 4 + 3 = 20\). Step 3: Calculate the mean: \(25 \div 20 = 1.25\).

M1: For a correct method to find the total number of goals (e.g. showing at least 2 correct products summed). M1: For dividing their total goals by the total frequency 20. A0.5: Correct answer of 1.25.

Calculate the numerator: \(\sqrt{8.41} = 2.9\) and \(3.2^2 = 10.24\). Sum of numerator = \(2.9 + 10.24 = 13.14\). Divide by the denominator: \(13.14 \div 0.5 = 26.28\).

M1: For evaluating \(\sqrt{8.41} = 2.9\) or \(3.2^2 = 10.24\). M1: For correctly finding the numerator sum as 13.14. A0.5: Correct final answer of 26.28.