2024 Edexcel GCSE Mathematics (1MA1) 模擬試題連答案詳解

To work out \( 0.3 \times 0.06 \), first calculate \( 3 \times 6 = 18 \). Since there are three decimal places in total in the numbers being multiplied, the final answer is \( 0.018 \).

B1 for 0.018 (or equivalent fraction, e.g. \(\frac{9}{500}\))

Multiply the coefficients: \( 5 \times 3 = 15 \). Then multiply the variables: \( a \times b = ab \). Combine these to get \( 15ab \).

B1 for 15ab (or equivalent, such as 15ba)

Find the highest common factor of 18 and 30, which is 6. Divide both numbers in the ratio by 6: \( 18 \div 6 = 3 \) and \( 30 \div 6 = 5 \). The simplest form is \( 3:5 \).

B1 for 3:5 (or 3 to 5)

Since \( 1 \text{ litre} = 1000 \text{ millilitres} \), multiply \( 4.2 \) by \( 1000 \): \( 4.2 \times 1000 = 4200 \text{ ml} \).

B1 for 4200 (allow 4200 ml)

The possible outcomes are 1, 2, 3, 4, 5, and 6. The prime numbers are 2, 3, and 5. There are 3 prime numbers. The probability is \( \frac{3}{6} \), which simplifies to \( \frac{1}{2} \).

B1 for 1/2

The range is the difference between the largest number and the smallest number. The largest number is 15 and the smallest number is 3. \( 15 - 3 = 12 \).

B1 for 12

To write \( 0.00073 \) in standard form, move the decimal point 4 places to the right to obtain \( 7.3 \). This gives a power of \( 10^{-4} \). The standard form is \( 7.3 \times 10^{-4} \).

B1 for 7.3 x 10^-4 (accept \( 7.3 \times 10^{-4} \))

Using the power rule of indices, multiply the exponents: \( 4 \times 5 = 20 \). Therefore, \( (y^4)^5 = y^{20} \).

B1 for y^20

To write a number in standard form, it must be in the form \( a \times 10^n \) where \( 1 \le a < 10 \) and \( n \) is an integer. For \( 0.00045 \), we move the decimal point 4 places to the right to get \( 4.5 \). Since we moved the decimal point to the right, the index is negative: \( -4 \). Therefore, the number in standard form is \( 4.5 \times 10^{-4} \).

B1 for \( 4.5 \times 10^{-4} \)

When multiplying terms with the same base, we add the indices: \( y^5 \times y^3 = y^{5 + 3} = y^8 \).

B1 for \( y^8 \)

First, convert the mixed numbers to improper fractions:
\( 1 \frac{3}{4} = \frac{7}{4} \)
\( 2 \frac{5}{8} = \frac{21}{8} \)

Next, perform the division of the two fractions:
\( \frac{7}{4} \div \frac{21}{8} = \frac{7}{4} \times \frac{8}{21} \)

Simplify the multiplication before multiplying out:
\( \frac{7 \times 8}{4 \times 21} = \frac{1 \times 2}{1 \times 3} = \frac{2}{3} \)

Finally, add \( \frac{1}{6} \) to the result:
\( \frac{2}{3} + \frac{1}{6} = \frac{4}{6} + \frac{1}{6} = \frac{5}{6} \)

M1: Convert both mixed numbers correctly to improper fractions: \( \frac{7}{4} \) and \( \frac{21}{8} \)
M1: Apply the correct method for division, e.g., \( \frac{7}{4} \times \frac{8}{21} \), to obtain \( \frac{2}{3} \) (or equivalent)
A1: \( \frac{5}{6} \) oe

Let \( A \) be the number of apples, \( B \) be the number of bananas, and \( O \) be the number of oranges.
We are given:
\( A : B = 2 : 3 \)
\( B : O = 4 : 5 \)

To combine these ratios, find a common multiple for the term representing bananas (\( B \)), which is the lowest common multiple of 3 and 4, i.e., 12.

Scale the first ratio \( A : B \) by multiplying by 4:
\( A : B = 8 : 12 \)

Scale the second ratio \( B : O \) by multiplying by 3:
\( B : O = 12 : 15 \)

Combining these gives the ratio:
\( A : B : O = 8 : 12 : 15 \)

Now, calculate the total number of parts in the ratio:
\( 8 + 12 + 15 = 35 \)

The fraction of bananas in the fruit bowl is:
\( \frac{12}{35} \)

M1: Find a common ratio scaling for B (e.g., scaling \( 2:3 \) to \( 8:12 \) or \( 4:5 \) to \( 12:15 \))
M1: Obtain a combined ratio for all three fruits: \( 8 : 12 : 15 \) (or equivalent ratio) or find the total parts of 35
A1: \( \frac{12}{35} \) oe

Since the two equal sides are \( 2x + 3 \) and \( 3x - 1 \), we can set up the equation:
\( 2x + 3 = 3x - 1 \)

Subtract \( 2x \) from both sides:
\( 3 = x - 1 \)

Add 1 to both sides:
\( x = 4 \)

Now, substitute \( x = 4 \) to find the length of each side:
Side 1: \( 2(4) + 3 = 11 \) cm
Side 2: \( 3(4) - 1 = 11 \) cm
Side 3: \( 4 + 5 = 9 \) cm

Calculate the perimeter by adding all three sides together:
Perimeter = \( 11 + 11 + 9 = 31 \) cm

M1: Set up the correct algebraic equation: \( 2x + 3 = 3x - 1 \)
M1: Solve to find \( x = 4 \) and substitute into at least one expression to find a side length (e.g., 11 or 9)
A1: 31 (accept 31 cm)

First, calculate the area of the rectangle:
Area of rectangle = \( 12 \times 5 = 60 \text{ cm}^2 \)

Next, calculate the area of the trapezium:
Area of trapezium = \( 1.2 \times 60 = 72 \text{ cm}^2 \)

The formula for the area of a trapezium is \( \frac{1}{2}(a + b)h \):
\( \frac{1}{2}(6 + 10)h = 72 \)
\( \frac{1}{2}(16)h = 72 \)
\( 8h = 72 \)

Solve for \( h \):
\( h = \frac{72}{8} = 9 \)

M1: Calculate the area of the rectangle: \( 12 \times 5 = 60 \) (or write an expression for the trapezium's area, e.g. \( 8h \))
M1: Form the equation \( 8h = 1.2 \times 60 \) or \( 8h = 72 \)
A1: 9

The sum of all probabilities is 1.
Therefore, the probability of selecting either a blue or a green counter is:
\( 1 - 0.35 = 0.65 \)

This probability is divided between blue and green counters in the ratio of \( 2 : 3 \).
Total parts in the ratio = \( 2 + 3 = 5 \) parts.

Calculate the probability of selecting a green counter:
\( P(\text{Green}) = \frac{3}{5} \times 0.65 \)
\( P(\text{Green}) = 3 \times 0.13 = 0.39 \)

M1: Subtract 0.35 from 1 to find the combined probability of blue and green counters: \( 1 - 0.35 = 0.65 \)
M1: Multiply the combined probability by \( \frac{3}{5} \) (e.g., \( 0.65 \div 5 \times 3 \))
A1: 0.39 (or equivalent fraction, e.g., \( \frac{39}{100} \))

First, find the total sum of the original 5 numbers:
\( \text{Sum of 5 numbers} = 5 \times 8 = 40 \)

Next, find the total sum of the 6 numbers including the new number:
\( \text{Sum of 6 numbers} = 6 \times 9.5 = 57 \)

The value of the 6th number is the difference between these two sums:
\( \text{6th number} = 57 - 40 = 17 \)

M1: Work out the sum of the original 5 numbers: \( 5 \times 8 = 40 \)
M1: Work out the sum of the 6 numbers: \( 6 \times 9.5 = 57 \)
A1: 17

Let \( x = 0.2363636... \)

Multiply by 10 to shift the non-repeating part past the decimal point:
\( 10x = 2.363636... \) [Equation 1]

Multiply by 1000 to shift one full repeating cycle past the decimal point:
\( 1000x = 236.363636... \) [Equation 2]

Subtract Equation 1 from Equation 2:
\( 1000x - 10x = 236.363636... - 2.363636... \)
\( 990x = 234 \)

Solve for \( x \):
\( x = \frac{234}{990} \)

Simplify the fraction by dividing the numerator and denominator by 9:
\( x = \frac{26}{110} \)

Simplify further by dividing by 2:
\( x = \frac{13}{55} \)

M1: Set up two equations to align the recurring decimal parts, e.g., \( 10x = 2.3636... \) and \( 1000x = 236.3636... \) (or equivalent)
M1: Subtract equations to obtain a correct linear relation, e.g., \( 990x = 234 \) (or \( 99x = 23.4 \))
A1: \( \frac{13}{55} \) (must be in simplest form)

Factorise the numerator: \( 2x^2 - 5x - 3 \)
We look for two numbers that multiply to \( 2 \times (-3) = -6 \) and add up to \( -5 \). These numbers are \( -6 \) and \( 1 \).
\( 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3) \)

Factorise the denominator: \( x^2 - 9 \)
This is a difference of two squares:
\( x^2 - 9 = (x - 3)(x + 3) \)

Write the fraction with the factorised expressions and cancel the common factor \( (x - 3) \):
\( \frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} = \frac{2x + 1}{x + 3} \)

M1: Factorise the denominator correctly as \( (x - 3)(x + 3) \)
M1: Factorise the numerator correctly as \( (2x + 1)(x - 3) \)
A1: \( \frac{2x + 1}{x + 3} \) (accept \( (2x + 1)/(x + 3) \))

Let the total number of books be \(x\).

1. The fraction of fiction books is \(\frac{3}{8}\), so the remaining fraction is \(1 - \frac{3}{8} = \frac{5}{8}\).

2. \(40\%\) of the remaining books are non-fiction, which is \(0.40 \times \frac{5}{8}x = \frac{2}{5} \times \frac{5}{8}x = \frac{1}{4}x\).

3. The reference books represent the remaining fraction of these books: \(100\% - 40\% = 60\%\) of the remaining books. So, Reference books = \(0.60 \times \frac{5}{8}x = \frac{3}{5} \times \frac{5}{8}x = \frac{3}{8}x\).

4. We are given that there are \(45\) reference books: \(\frac{3}{8}x = 45\).

5. Solve for \(x\): \(x = 45 \times \frac{8}{3} = 15 \times 8 = 120\).

M1: For finding the remaining fraction \(\frac{5}{8}\) or for finding \(60\%\) of the remaining books, e.g. \(0.6 \times \frac{5}{8}\)
M1: For setting up the equation \(\frac{3}{8}x = 45\) or an equivalent proportion (e.g. 3 parts = 45)
A1: 120

1. Write down an expression for the perimeter of the rectangle:
\(P_{\text{rect}} = 2((2x + 5) + (x + 1)) = 2(3x + 6) = 6x + 12\).

2. Write down an expression for the perimeter of the equilateral triangle:
\(P_{\text{tri}} = 3(x + 7) = 3x + 21\).

3. Set the two perimeters equal to each other:
\(6x + 12 = 3x + 21\).

4. Solve for \(x\):
\(3x = 9\)
\(x = 3\).

M1: For a correct expression for the perimeter of the rectangle, e.g. \(2(2x + 5 + x + 1)\) or \(6x + 12\), or for the triangle, e.g. \(3(x + 7)\) or \(3x + 21\)
M1: For equating the two perimeter expressions and starting to solve, e.g. \(6x + 12 = 3x + 21\) leading to \(3x = 9\)
A1: 3

1. Combine the two ratios by finding a common term for the blue counters (LCM of 4 and 5 is 20):
Red : Blue = \(3 \times 5 : 4 \times 5 = 15 : 20\)
Blue : Yellow = \(5 \times 4 : 2 \times 4 = 20 : 8\)

2. The combined ratio of Red : Blue : Yellow is \(15 : 20 : 8\).

3. Let the number of parts for red, blue, and yellow counters be \(15p\), \(20p\), and \(8p\) respectively.
The difference between red and yellow counters is:
\(15p - 8p = 7p\).

4. We are given that there are \(49\) more red than yellow counters, so:
\(7p = 49 \implies p = 7\).

5. The total number of counters is:
\(15p + 20p + 8p = 43p\).
Total = \(43 \times 7 = 301\).

M1: For finding a combined ratio Red : Blue : Yellow as \(15 : 20 : 8\) (or equivalent)
M1: For setting up an equation for the difference, e.g. \(15p - 8p = 49\) or \(7 \text{ parts} = 49\), and finding the value of 1 part as 7
A1: 301

1. Write down the formula for the area of the trapezium cross-section:
\(\text{Area} = \frac{1}{2}(a+b)h = \frac{1}{2}(6 + 10)h = 8h\).

2. Use the volume of a prism formula (Volume = Area of cross-section \(\times\) length):
\(\text{Volume} = 8h \times 12 = 96h\).

3. Set the volume equal to \(384\):
\(96h = 384\).

4. Solve for \(h\):
\(h = \frac{384}{96} = 4\).

M1: For a correct expression for the area of the cross-section, e.g., \(\frac{1}{2}(6+10)h\) or \(8h\)
M1: For setting up the volume equation, e.g., \(8h \times 12 = 384\) or \(96h = 384\)
A1: 4

1. Let the probability of choosing a green counter be \(x\). Then the probability of choosing a blue counter is \(2x\).

2. The sum of the probabilities must equal 1:
\(0.3 + 2x + x = 1\)
\(0.3 + 3x = 1\)
\(3x = 0.7\)
\(x = \frac{0.7}{3} = \frac{7}{30}\).
So, the probability of choosing a green counter is \(\frac{7}{30}\).

3. Let the total number of counters be \(T\). Since there are 14 green counters, we have:
\(\frac{7}{30} \times T = 14\)
\(T = 14 \times \frac{30}{7} = 2 \times 30 = 60\).

M1: For forming an equation for probabilities, e.g., \(0.3 + 3x = 1\), or finding the combined probability of blue and green as 0.7
M1: For finding the probability of green as \(\frac{7}{30}\) (or 0.233...) or establishing that green represents \(\frac{7}{30}\) of the total
A1: 60

Let the five integers in ascending order be \(a, b, c, d, e\).

1. Since the median is 8, the middle value is 8: \(c = 8\).
2. The mode is 6. Since 6 < 8, both \(a\) and \(b\) must be 6: \(a = 6, b = 6\).
3. The range is 10. Range = \(e - a = 10\). Since \(a = 6\), then \(e = 16\).
4. The mean is 9. The sum of the 5 numbers is \(5 \times 9 = 45\).
Sum = \(6 + 6 + 8 + d + 16 = 36 + d = 45\).
Therefore, \(d = 9\).

The five integers are 6, 6, 8, 9, 16.

M1: For using the median to find the middle number is 8 AND the mode to find the first two numbers are 6 (i.e. finding 6, 6, 8)
M1: For using the range to find the largest number is 16 (6 + 10) OR setting up the sum equation \(6 + 6 + 8 + d + 16 = 45\)
A1: 6, 6, 8, 9, 16 (any order, but must be these exact five numbers)

For part (a): Since the total volume of \(63\pi\text{ cm}^3\) is divided in the ratio \(2 : 5\) between the hemisphere and the cylinder, we first find the shares. Total shares = \(2 + 5 = 7\). One share is \(\frac{63\pi}{7} = 9\pi\text{ cm}^3\). Therefore, Volume of hemisphere = \(2 \times 9\pi = 18\pi\text{ cm}^3\) and Volume of cylinder = \(5 \times 9\pi = 45\pi\text{ cm}^3\). Using the formula for the volume of a hemisphere, \(V = \frac{2}{3}\pi r^3\): \(\frac{2}{3}\pi r^3 = 18\pi \implies \frac{2}{3}r^3 = 18 \implies r^3 = 27 \implies r = 3\). Using the formula for the volume of a cylinder, \(V = \pi r^2 h\): \(\pi (3^2) h = 45\pi \implies 9\pi h = 45\pi \implies h = 5\). For part (b): Mass = Density \(\times\) Volume. Total mass = \(8 \times 63\pi = 504\pi\text{ g}\).

Part (a): M1 for dividing \(63\pi\) into the ratio \(2 : 5\) to find either \(18\pi\) or \(45\pi\). M1 for setting up \(\frac{2}{3}\pi r^3 = 18\pi\) (or equivalent). A1 for \(r = 3\). M1 for substituting their \(r\) into the cylinder volume formula: \(\pi \times 3^2 \times h = 45\pi\) (or equivalent). A1 for \(h = 5\). Part (b): M1 for calculating \(\text{mass} = 8 \times 63\pi\). A1 for \(504\pi\) (accept with or without unit \(g\)).

For part (a): Area of rectangle = \((2x - 3)(x + 4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12\). Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(x + 2)(2x + 8) = (x + 2)(x + 4) = x^2 + 6x + 8\). Since the areas are equal: \(2x^2 + 5x - 12 = x^2 + 6x + 8\). Rearranging gives: \(2x^2 - x^2 + 5x - 6x - 12 - 8 = 0 \implies x^2 - x - 20 = 0\). For part (b): Solving \(x^2 - x - 20 = 0 \implies (x - 5)(x + 4) = 0\), which gives \(x = 5\) or \(x = -4\). We must reject \(x = -4\) because it results in a negative width for the rectangle: \(2(-4) - 3 = -11\) cm, and lengths must be positive. Therefore, \(x = 5\). For part (c): Substituting \(x = 5\), the rectangle width is \(2(5) - 3 = 7\) cm and the height is \(5 + 4 = 9\) cm. Perimeter of rectangle = \(2(7 + 9) = 32\) cm. Perimeter of square = \(32\) cm. Side length of square = \(\frac{32}{4} = 8\) cm. Area of square = \(8^2 = 64\text{ cm}^2\).

Part (a): M1 for expanding rectangle dimensions to get \(2x^2 + 5x - 12\) (allow one sign error). M1 for setting up the area of the triangle and simplifying to \(x^2 + 6x + 8\). A1 for equating and completing the proof to show \(x^2 - x - 20 = 0\) with no algebraic errors. Part (b): M1 for factorising to \((x - 5)(x + 4) = 0\) or identifying roots as \(5\) and \(-4\). A1 for selecting \(x = 5\) and explaining that \(x = -4\) leads to negative lengths. Part (c): M1 for finding the perimeter of the rectangle as \(32\) cm. A1 for correct area of the square as \(64\) (accept with or without unit \(\text{cm}^2\)).

Let the original number of red counters be \(3y\) and blue counters be \(5y\). This means the sum of red and blue counters is \(8y\). Let the original number of green counters be \(g\). The total number of counters is \(8y + g\). The probability of picking a green counter is \(\frac{g}{8y + g} = \frac{1}{5}\). Multiplying both sides gives \(5g = 8y + g \implies 4g = 8y \implies g = 2y\). So, the number of green counters is \(2y\), and the total number of counters initially is \(10y\). When \(2\) green counters are added: New number of green counters is \(2y + 2\). New total number of counters is \(10y + 2\). The new probability is \(\frac{2y+2}{10y+2} = \frac{1}{4}\). Cross-multiplying: \(4(2y + 2) = 10y + 2 \implies 8y + 8 = 10y + 2 \implies 2y = 6 \implies y = 3\). Substituting \(y = 3\) back into our definitions: Red counters = \(3(3) = 9\). Blue counters = \(5(3) = 15\). Green counters = \(2(3) = 6\).

M1 for expressing red as \(3y\) and blue as \(5y\) (or total red and blue as \(8y\)). M1 for setting up the equation \(\frac{g}{8y+g} = \frac{1}{5}\) (or equivalent). M1 for finding the relation \(g = 2y\) or expressing green as a fraction of the other counters. M1 for setting up the new probability fraction with the addition of 2 green counters: \(\frac{2y+2}{10y+2} = \frac{1}{4}\). M1 for algebraic manipulation to solve for their variable (e.g., cross-multiplying). A1 for finding the parameter \(y = 3\) or the original number of green counters as \(6\). A1 for the correct original numbers: \(9\) red, \(15\) blue, \(6\) green (must be clearly identified).

For part (a): Gradient of \(L_1 = \frac{8 - 0}{6 - 2} = \frac{8}{4} = 2\). Using \(y - y_1 = m(x - x_1)\) with point \(A(2,0)\): \(y - 0 = 2(x - 2) \implies y = 2x - 4\). For part (b): Midpoint of \(AB = \left(\frac{2+6}{2}, \frac{0+8}{2}\right) = (4, 4)\). The gradient of the perpendicular line \(L_2\) is \(-\frac{1}{2}\). Using \(y - y_1 = m_2(x - x_1)\) with midpoint \((4,4)\): \(y - 4 = -\frac{1}{2}(x - 4)\). Multiply by 2: \(2y - 8 = -x + 4\). Rearrange to the required form: \(x + 2y - 12 = 0\). For part (c): \(P\) is on the x-axis, so \(y = 0 \implies x + 2(0) - 12 = 0 \implies x = 12\). So \(P = (12, 0)\). \(Q\) is on the y-axis, so \(x = 0 \implies 0 + 2y - 12 = 0 \implies y = 6\). So \(Q = (0, 6)\). The area of triangle \(OPQ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 6 = 36\).

Part (a): M1 for finding the gradient as \(2\). A1 for \(y = 2x - 4\). Part (b): M1 for finding the midpoint of \(AB\) as \((4, 4)\) or finding perpendicular gradient \(-\frac{1}{2}\). M1 for substituting their midpoint and perpendicular gradient into \(y - y_1 = m(x - x_1)\) or equivalent. A1 for \(x + 2y - 12 = 0\) (or any integer multiple thereof). Part (c): M1 for finding the intercepts \(P(12, 0)\) and \(Q(0, 6)\) from their \(L_2\) equation. A1 for correct area of \(36\).

The reciprocal of a number \(x\) is \(\frac{1}{x}\). The reciprocal of 2.5 is \(\frac{1}{2.5} = 0.4\).

B1 for 0.4 (or equivalent fraction such as \(\frac{2}{5}\))

Using the laws of indices, \(y^a \div y^b = y^{a-b}\). Therefore, \(y^6 \div y^2 = y^{6-2} = y^4\).

B1 for \(y^4\)

Divide both parts of the ratio by their highest common factor, which is 5. \(15 \div 5 = 3\) and \(35 \div 5 = 7\). So the ratio in its simplest form is 3 : 7.

B1 for 3:7 (or 3 : 7)

A 3D shape with 6 faces where all faces are identical squares is a cube.

B1 for cube

The outcomes on a fair 6-sided dice are 1, 2, 3, 4, 5, 6. The prime numbers are 2, 3, and 5. There are 3 prime numbers out of 6 possible outcomes. Probability = \(\frac{3}{6} = 0.5\).

B1 for 0.5 or \(\frac{1}{2}\) (or equivalent fraction)

The mode is the value that appears most frequently in the list. The number 8 appears twice, while all other numbers appear once. So the mode is 8.

B1 for 8

To convert metres per second to kilometres per hour: \(15\text{ m/s} = 15 \times 60\text{ m/min} = 900\text{ m/min}\). \(900 \times 60\text{ m/hour} = 54000\text{ m/hour}\). Divide by 1000 to convert metres to kilometres: \(54000 \div 1000 = 54\text{ km/h}\).

B1 for 54

To find the value of \(64^{\frac{2}{3}}\), first find the cube root of 64, then square the result. \(\sqrt[3]{64} = 4\). \(4^2 = 16\).

B1 for 16

First, calculate the value of the investment at the end of the second year: \(4500 \times 1.023^2 = 4709.3805\) pounds. Next, calculate the value of the investment at the end of the third year: \(4709.3805 \times 1.015 = 4780.0212075\) pounds. Now, find the total interest earned by subtracting the initial investment: \(4780.0212075 - 4500 = 280.0212075\) pounds. Rounding to the nearest penny gives £280.02.

M1 for calculating the amount after 2 years: \(4500 \times 1.023^2\) or \(4709.38\). M1 for completing the compound interest calculation: \(4709.38 \times 1.015\) (or \(4780.02\)). A1 for 280.02 (accept £280.02).

Perimeter of the rectangle is \(2(x + 3 + 3x - 1) = 2(4x + 2) = 8x + 4\) cm. Perimeter of the equilateral triangle is \(3(2x + 4) = 6x + 12\) cm. Since the perimeters are equal, we can write: \(8x + 4 = 6x + 12\). Subtracting \(6x\) from both sides gives \(2x + 4 = 12\). Subtracting 4 from both sides gives \(2x = 8\), which simplifies to \(x = 4\).

M1 for writing a correct expression for the perimeter of the rectangle (e.g. \(8x + 4\)) or the triangle (e.g. \(6x + 12\)). M1 for equating the two perimeters and forming a correct equation: \(8x + 4 = 6x + 12\). A1 for \(x = 4\).

The difference between C's share and A's share is \(8 - 3 = 5\) parts. We are given that this difference is £120, so \(5\text{ parts} = 120\). This means \(1\text{ part} = 120 \div 5 = 24\) pounds. The total number of parts shared is \(3 + 5 + 8 = 16\) parts. The total amount of money shared is \(16 \times 24 = 384\) pounds.

M1 for finding the difference in parts between A and C: \(8 - 3 = 5\). M1 for calculating the value of one part: \(120 \div 5 = 24\). A1 for 384.

First, find the area of the trapezium cross-section: \(\text{Area} = \frac{1}{2}(8 + 12) \times 5 = 50\text{ cm}^2\). Next, calculate the volume of the prism: \(\text{Volume} = 50 \times 15 = 750\text{ cm}^3\). Finally, calculate the mass of the prism using its density: \(\text{Mass} = 750 \times 0.8 = 600\text{ g}\).

M1 for calculating the area of the cross-section: \(\frac{1}{2}(8+12) \times 5 = 50\). M1 for finding the volume of the prism (\(50 \times 15 = 750\)) and multiplying by the density (\(0.8\)). A1 for 600.

The total probability of blue and yellow counters is \(1 - 0.2 = 0.8\). The ratio of blue to yellow is \(5 : 3\), so the probability of picking a yellow counter is \(\frac{3}{8} \times 0.8 = 0.3\). Since the probability of a yellow counter is \(0.3\) and there are \(15\) yellow counters, we set up the equation: \(\frac{15}{\text{Total}} = 0.3\). Thus, \(\text{Total} = \frac{15}{0.3} = 50\).

M1 for finding the combined probability of blue and yellow counters: \(1 - 0.2 = 0.8\). M1 for finding the probability of a yellow counter: \(\frac{3}{8} \times 0.8 = 0.3\) (or equivalent method). A1 for 50.

Calculate the total height of the boys: \(12 \times 1.62 = 19.44\text{ m}\). Calculate the total height of the girls: \(8 \times 1.54 = 12.32\text{ m}\). The overall total height for all 20 children is \(19.44 + 12.32 = 31.76\text{ m}\). The mean height is \(31.76 \div 20 = 1.588\text{ m}\).

M1 for calculating the total height of either group: \(12 \times 1.62 = 19.44\) or \(8 \times 1.54 = 12.32\). M1 for adding both total heights and dividing by 20: \((19.44 + 12.32) \div 20\). A1 for 1.588.

In triangle \(ABC\), the hypotenuse \(AC\) can be found using cosine: \(\cos(35^\circ) = \frac{7.2}{AC}\), which gives \(AC = \frac{7.2}{\cos(35^\circ)} \approx 8.7896\text{ cm}\). In triangle \(ACD\), \(AC\) is the hypotenuse, so by Pythagoras' theorem: \(AC^2 = AD^2 + CD^2\). Rearranging gives \(AD^2 = AC^2 - CD^2 = 8.7896^2 - 5.4^2 \approx 77.2571 - 29.16 = 48.0971\). Therefore, \(AD = \sqrt{48.0971} \approx 6.9352\text{ cm}\). To 3 significant figures, this is \(6.94\text{ cm}\).

M1 for finding the length of \(AC\): \(AC = \frac{7.2}{\cos(35)}\) (or \(8.79\)). M1 for applying Pythagoras' theorem in triangle \(ACD\): \(AD = \sqrt{AC^2 - 5.4^2}\). A1 for 6.94 (accept 6.93 to 6.95).

Since \(y\) is inversely proportional to \(x^2\), we can write the equation: \(y = \frac{k}{x^2}\). Substitute \(x = 5\) and \(y = 12\) to find \(k\): \(12 = \frac{k}{5^2} \implies k = 12 \times 25 = 300\). When \(x = 2.5\), calculate \(y\): \(y = \frac{300}{2.5^2} = \frac{300}{6.25} = 48\).

M1 for setting up the proportional relationship \(y = \frac{k}{x^2}\). M1 for finding the constant of proportionality \(k = 300\). A1 for 48.

To find a combined ratio for Red (R), Blue (B), and Green (G), we find a common multiple for B. The ratio R : B is 3 : 5, and B : G is 4 : 3. The lowest common multiple of 5 and 4 is 20. Multiplying R : B by 4 gives 12 : 20. Multiplying B : G by 5 gives 20 : 15. Thus, the combined ratio R : B : G is 12 : 20 : 15. The difference in parts between blue and green counters is \(20 - 15 = 5\) parts. We are given that this difference represents 25 counters. Therefore, 1 part represents \(25 \div 5 = 5\) counters. The total number of parts is \(12 + 20 + 15 = 47\) parts. The total number of counters is \(47 \times 5 = 235\).

M1: For a process to find a combined ratio for R : B : G, e.g., finding a common multiple for B to get \(12 : 20 : 15\). M1: For a process to find the value of one share, e.g., \(25 \div (20 - 15)\). A1: For 235.

First, calculate the area of the rectangular lawn: \(\text{Area} = 12 \times 8 = 96\text{ m}^2\). Next, calculate the area of the circular flower bed. The radius is \(3.5 \div 2 = 1.75\text{ m}\). \(\text{Area of circle} = \pi \times 1.75^2 \approx 9.621\text{ m}^2\). The remaining grass area is \(96 - 9.621 = 86.379\text{ m}^2\). Find the number of boxes of seed required: \(86.379 \div 6 \approx 14.397\) boxes. Since we must buy whole boxes, we round up to 15 boxes. Finally, calculate the total cost: \(15 \times £4.50 = £67.50\).

M1: For a complete method to find the remaining grass area, e.g., \(12 \times 8 - \pi \times 1.75^2\) (yielding a value in the range 86.3 to 86.4). M1: For dividing their grass area by 6 and rounding up to the next integer to find the number of boxes (15 boxes). A1: For £67.50.

The perimeter of a rectangle is the sum of all four sides: \(2(3x + 5) + 2(2x - 1) = 48\). Expanding the brackets gives \(6x + 10 + 4x - 2 = 48\). Simplifying gives \(10x + 8 = 48\). Subtracting 8 from both sides gives \(10x = 40\), so \(x = 4\). Now, substitute \(x = 4\) to find the dimensions of the rectangle: \(\text{Length} = 3(4) + 5 = 17\text{ cm}\), and \(\text{Width} = 2(4) - 1 = 7\text{ cm}\). The area is \(\text{Length} \times \text{Width} = 17 \times 7 = 119\text{ cm}^2\).

M1: For setting up and simplifying a perimeter equation, e.g., \(2(3x + 5) + 2(2x - 1) = 48\) leading to \(10x + 8 = 48\). M1: For solving to find \(x = 4\) and substituting to find length (17) and width (7). A1: For 119.

The total number of students studying at least one of the two subjects is \(80 - 12 = 68\). Let \(H\) be the set of students studying History and \(G\) be the set of students studying Geography. We use the formula \(n(H \cup G) = n(H) + n(G) - n(H \cap G)\). Substituting the known values: \(68 = 45 + 40 - n(H \cap G)\), which simplifies to \(68 = 85 - n(H \cap G)\). Therefore, \(n(H \cap G) = 85 - 68 = 17\) students study both. The probability of choosing a student who studies both is \(\frac{17}{80}\) (or 0.2125).

M1: For subtracting the students who study neither from the total: \(80 - 12 = 68\). M1: For calculating the number of students who study both: \(45 + 40 - 68 = 17\). A1: For \(\frac{17}{80}\) or \(0.2125\).

Let \(V\) be the initial value of the company at the start of 2021. The multiplier for a 12% increase is 1.12. The multiplier for a 5% decrease is 0.95. The equation representing the changes is \(V \times 1.12 \times 0.95 = 319200\). Combining the multipliers gives \(1.12 \times 0.95 = 1.064\). So, \(1.064V = 319200\). Solving for \(V\) gives \(V = 319200 \div 1.064 = 300000\).

M1: For identifying both multipliers as 1.12 and 0.95, or combining them to get 1.064. M1: For setting up the reverse percentage process, e.g., \(319200 \div 1.064\). A1: For 300000.

First, find the midpoint of each interval: 135, 145, 155, and 165. Next, multiply each midpoint by its corresponding frequency: \(135 \times 4 = 540\), \(145 \times 9 = 1305\), \(155 \times 8 = 1240\), and \(165 \times 4 = 660\). Sum these products to find the total estimated height: \(540 + 1305 + 1240 + 660 = 3745\). Divide this total by the sum of the frequencies (25) to find the estimated mean: \(3745 \div 25 = 149.8\) cm.

M1: For finding the midpoints of the intervals (at least three correct) or showing the products of midpoints and frequencies. M1: For dividing the sum of their products \(\sum f x\) by the sum of frequencies (25). A1: For 149.8.

First, calculate the volume of water in the tank: \(V = \pi \times r^2 \times h = \pi \times 1.2^2 \times 1.5 = 2.16\pi \approx 6.78584\text{ m}^3\). Next, convert this volume to litres: \(6.78584 \times 1000 \approx 6785.84\text{ litres}\). Calculate the total time in minutes: \(4\text{ hours} = 4 \times 60 = 240\text{ minutes}\). Find the rate of flow in litres per minute: \(6785.84 \div 240 \approx 28.274\text{ litres/minute}\). To 3 significant figures, this is 28.3.

M1: For calculating the volume of water in \(\text{m}^3\), e.g., \(\pi \times 1.2^2 \times 1.5\) (approx. 6.79). M1: For converting volume to litres and dividing by the total minutes (240). A1: For 28.3 (accept 28.2 to 28.3).

Multiply both sides by \(x - 4\) to clear the fraction: \(y(x - 4) = 3x + 2\). Expand the brackets: \(xy - 4y = 3x + 2\). Group all terms containing \(x\) on one side and the other terms on the opposite side: \(xy - 3x = 4y + 2\). Factorise \(x\) from the left side: \(x(y - 3) = 4y + 2\). Finally, divide by \(y - 3\) to isolate \(x\): \(x = \frac{4y + 2}{y - 3}\).

M1: For multiplying by \(x-4\) and expanding the left-hand side: \(xy - 4y = 3x + 2\). M1: For isolating terms in \(x\) on one side and factorising: \(x(y - 3) = 4y + 2\). A1: For \(x = \frac{4y + 2}{y - 3}\) (or equivalent expression, e.g., \(x = \frac{-4y - 2}{3 - y}\)).

1. Volume of one cylinder: \(V = \pi r^2 h = \pi \times 3^2 \times 8 = 72\pi \approx 226.1947\text{ cm}^3\). 2. Total volume of 250 cylinders: \(250 \times 72\pi = 18000\pi \approx 56548.6678\text{ cm}^3\). 3. Total mass of 250 cylinders: \(\text{Mass} = \text{Volume} \times \text{Density} = 56548.6678 \times 7.8 = 441079.6087\text{ g}\). 4. Convert mass to kg: \(441.0796087\text{ kg}\). 5. Total mass to order including 5% waste: \(441.0796087 \times 1.05 = 463.133589\text{ kg}\). 6. Total cost: \(463.133589 \times 4.20 = 1945.16107\) which rounds to \(£1945.16\).

M1: for attempting to find the volume of one cylinder: \(\pi \times 3^2 \times 8\) or \(72\pi\). M1: for multiplying by 250 and then by 7.8 to find the total mass in grams (approx. 441080 g). M1: for dividing the mass by 1000 to convert to kg (approx. 441.1 kg). M1: for accounting for the 5% waste by multiplying by 1.05 (approx. 463.1 kg). M1: for multiplying the mass in kg by 4.20. A1: for the correct final cost of £1945.16 (accept £1945.16 to £1945.17 depending on rounding methods).

1. Area of the outer rectangle: \((2x + 5)(x + 3) = 2x^2 + 11x + 15\). 2. Area of the square flower bed: \((x - 1)^2 = x^2 - 2x + 1\). 3. Set up the equation for the remaining area: \((2x^2 + 11x + 15) - (x^2 - 2x + 1) = 115\). Simplifying this gives \(x^2 + 13x + 14 = 115\), which rearranges to \(x^2 + 13x - 101 = 0\). 4. Solve the quadratic equation using the quadratic formula: \(x = \frac{-13 \pm \sqrt{13^2 - 4(1)(-101)}}{2} = \frac{-13 \pm \sqrt{169 + 404}}{2} = \frac{-13 \pm \sqrt{573}}{2}\). Since \(x > 1\), we choose the positive root: \(x = \frac{-13 + 23.9374}{2} \approx 5.4687\text{ m}\). 5. Perimeter of the outer lawn: \(P = 2((2x + 5) + (x + 3)) = 2(3x + 8) = 6x + 16\). Substituting \(x \approx 5.4687\): \(P = 6(5.4687) + 16 \approx 48.8122\text{ m}\). To 3 significant figures, the perimeter is \(48.8\text{ m}\).

M1: for expansion of rectangular area as \(2x^2 + 11x + 15\). M1: for expansion of square area as \(x^2 - 2x + 1\). M1: for setting up the simplified quadratic equation \(x^2 + 13x - 101 = 0\). M1: for solving the quadratic equation to find \(x \approx 5.47\) (ignoring negative root). M1: for substituting their positive \(x\) into the expression for the perimeter, \(2(3x + 8)\) or \(6x + 16\). A1: for the correct perimeter of \(48.8\text{ m}\) (accept \(48.8\) or \(48.81\)).

1. Let the number of green counters be \(3x\) and red counters be \(4x\). The combined number of green and red counters is \(7x\). 2. Since the probability of choosing a blue counter is \(\frac{1}{5}\), the probability of choosing a non-blue counter (green or red) is \(\frac{4}{5}\). This means \(\frac{7x}{\text{Total}} = \frac{4}{5}\), so \(\text{Total} = \frac{35x}{4}\). 3. To work with integers, let \(x = 4k\). This gives: Green = \(12k\), Red = \(16k\), and Total = \(35k\). 4. The probability of picking two green counters without replacement is: \(P(G, G) = \frac{12k}{35k} \times \frac{12k - 1}{35k - 1} = \frac{12}{35} \times \frac{12k - 1}{35k - 1}\). 5. Set this equal to \(\frac{4}{35}\): \(\frac{12}{35} \times \frac{12k - 1}{35k - 1} = \frac{4}{35} \implies 3 \times \frac{12k - 1}{35k - 1} = 1 \implies 36k - 3 = 35k - 1 \implies k = 2\). 6. The total number of counters is \(35k = 35 \times 2 = 70\).

M1: for establishing algebraic expressions for Green, Red, and Blue counters in terms of a single variable, e.g., Green = \(12k\), Red = \(16k\), Blue = \(7k\), Total = \(35k\). M1: for writing the correct algebraic probability for selecting two green counters without replacement: \(\frac{12k}{35k} \times \frac{12k - 1}{35k - 1}\). M1: for setting up the equation: \(\frac{12}{35} \times \frac{12k - 1}{35k - 1} = \frac{4}{35}\). M1: for simplifying the equation to a linear form, e.g., \(3(12k - 1) = 35k - 1\). M1: for solving to find \(k = 2\) (or equivalent variable). A1: for the correct total number of counters of 70.

1. Determine the interior angle \(\angle ABC\). The bearing of \(B\) from \(A\) is \(075^\circ\), so the back-bearing of \(A\) from \(B\) is \(180^\circ + 75^\circ = 255^\circ\). The bearing of \(C\) from \(B\) is \(150^\circ\). Thus, \(\angle ABC = 255^\circ - 150^\circ = 105^\circ\). 2. Use the Cosine Rule to find distance \(AC\): \(AC^2 = 18^2 + 25^2 - 2 \times 18 \times 25 \times \cos(105^\circ) = 324 + 625 - 900(-0.258819) = 949 + 232.937 = 1181.937\). Hence, \(AC = \sqrt{1181.937} \approx 34.379\text{ km}\). 3. Use the Sine Rule to find the angle \(\angle BAC\): \(\frac{\sin(\angle BAC)}{25} = \frac{\sin(105^\circ)}{34.379} \implies \sin(\angle BAC) = \frac{25 \times \sin(105^\circ)}{34.379} \approx 0.7024\). Therefore, \(\angle BAC = \sin^{-1}(0.7024) \approx 44.62^\circ\). 4. Calculate the bearing of \(C\) from \(A\). The bearing of \(B\) from \(A\) is \(075^\circ\), so the bearing of \(C\) from \(A\) is \(075^\circ + 44.62^\circ = 119.62^\circ\). To the nearest degree, this is \(120^\circ\).

M1: for calculating the interior angle \(\angle ABC = 105^\circ\). M1: for substituting correctly into the cosine rule: \(AC^2 = 18^2 + 25^2 - 2(18)(25)\cos(105^\circ)\). A1: for finding the distance \(AC \approx 34.4\text{ km}\) (or showing \(34.379...\)). M1: for substituting correctly into the sine rule (or cosine rule) to find angle \(\angle BAC\): \(\frac{\sin(\angle BAC)}{25} = \frac{\sin(105^\circ)}{34.379}\). A1: for finding \(\angle BAC \approx 44.6^\circ\) (or \(45^\circ\)). A1: for the correct bearing of \(120^\circ\) (accept \(120\) or \(120^\circ\)).

In the number \(23.471\), the digit \(2\) is in the tens column, \(3\) is in the units column, \(4\) is in the tenths column, \(7\) is in the hundredths column, and \(1\) is in the thousandths column. Therefore, the value of the 7 is \(0.07\) (or seven hundredths).

B1 for \(0.07\) or \(\frac{7}{100}\) or seven hundredths.

To simplify \(3a \times 4b\), multiply the coefficients and the variables: \(3 \times 4 \times a \times b = 12ab\).

B1 for \(12ab\) (accept equivalent forms like \(12ba\)).

To simplify the ratio \(18 : 45\), find the highest common factor of \(18\) and \(45\), which is \(9\). Divide both numbers by \(9\): \(18 \div 9 = 2\) and \(45 \div 9 = 5\). So, the ratio in its simplest form is \(2 : 5\).

B1 for \(2 : 5\) (or \(2\text{ to }5\)).

Since \(1\text{ m} = 100\text{ cm}\), then \(1\text{ m}^3 = (100\text{ cm})^3 = 1,000,000\text{ cm}^3\). Therefore, \(4.5\text{ m}^3 = 4.5 \times 1,000,000 = 4,500,000\text{ cm}^3\).

B1 for \(4,500,000\) (or \(4.5 \times 10^6\)).

The possible outcomes on a six-sided dice are \(1, 2, 3, 4, 5, 6\). The prime numbers in this set are \(2, 3,\) and \(5\). There are \(3\) prime numbers out of \(6\) outcomes in total. The probability is \(\frac{3}{6} = \frac{1}{2} = 0.5\).

B1 for \(0.5\), \(\frac{1}{2}\) or \(50\%\) (accept any equivalent fraction).

First, arrange the numbers in ascending order: \(5, 7, 8, 9, 11, 12, 14\). There are 7 numbers, so the median is the middle number (the 4th number), which is \(9\).

B1 for \(9\).

To write \(0.000038\) in standard form, we move the decimal point 5 places to the right to get a number between 1 and 10, which is \(3.8\). Since we moved the decimal point to the right, the power of 10 is negative. Thus, the number is \(3.8 \times 10^{-5}\).

B1 for \(3.8 \times 10^{-5}\) (or equivalent).

To factorise \(6x^2 - 9x\), we find the highest common factor of the terms \(6x^2\) and \(-9x\). The highest common factor is \(3x\). Dividing each term by \(3x\) gives \(2x - 3\). So, the factorised expression is \(3x(2x - 3)\).

B1 for \(3x(2x - 3)\).

First, convert both quantities into the same units.
Since \(1\text{ km} = 1000\text{ m}\):
\(1.25\text{ km} = 1.25 \times 1000\text{ m} = 1250\text{ m}\).

Now, write the ratio of the two quantities:
\(250 : 1250\)

To write the ratio in the form \(1 : n\), divide both sides by \(250\):
\(\frac{250}{250} : \frac{1250}{250} = 1 : 5\)

Therefore, \(n = 5\) and the ratio is \(1 : 5\).

B1: For \(1 : 5\) (accept \(n = 5\))

First, convert both quantities into the same units.
Since \(1\text{ km} = 1000\text{ m}\):
\(1.25\text{ km} = 1.25 \times 1000\text{ m} = 1250\text{ m}\).

Now, write the ratio of the two quantities:
\(250 : 1250\)

To write the ratio in the form \(1 : n\), divide both sides by \(250\):
\(\frac{250}{250} : \frac{1250}{250} = 1 : 5\)

Therefore, \(n = 5\) and the ratio is \(1 : 5\).

B1: For \(1 : 5\) (accept \(n = 5\))

First, find the total number of parts in the ratio:
\(11 + 1 = 12\) parts.

Next, calculate the mass of one part:
\(36 \div 12 = 3\) grams.

Find the mass of gold and copper in the coin:
Mass of gold = \(11 \times 3 = 33\) grams.
Mass of copper = \(1 \times 3 = 3\) grams.

Calculate the cost of each metal:
Cost of gold = \(33 \times 48 = £1584\).
Cost of copper = \(3 \times 0.05 = £0.15\).

Calculate the total cost:
\(1584 + 0.15 = 1584.15\).

M1: for dividing the total mass by the sum of the ratio parts to find the unit mass, e.g. \(36 \div (11+1)\), or finding the correct mass of gold (33 g) and copper (3 g).
M1: for a complete method to calculate the total cost, e.g. \((33 \times 48) + (3 \times 0.05)\).
A1: for 1584.15 (accept £1584.15).

First, find the area of the rectangular garden:
\(\text{Area of rectangle} = 12 \times 8 = 96\text{ m}^2\).

Next, find the area of the circular pond:
\(\text{Area of circle} = \pi \times r^2 = \pi \times 2.5^2 = 6.25\pi \approx 19.635\text{ m}^2\).

Subtract the area of the pond from the area of the garden to find the grass area:
\(\text{Area of grass} = 96 - 19.635 = 76.365\text{ m}^2\).

Rounding to 3 significant figures gives \(76.4\text{ m}^2\).

M1: for finding the area of the rectangle \(12 \times 8 = 96\) or the area of the circular pond \(\pi \times 2.5^2\) (implied by \(19.6\) or \(6.25\pi\)).
M1: for a complete method of subtracting the area of the pond from the area of the rectangle, e.g. \(96 - \pi \times 2.5^2\).
A1: for 76.4 (accept 76.3 to 76.4).

Let \(x\) be the cost of one t-shirt and \(y\) be the postage fee.
From the given information, we can write two equations:
1) \(3x + y = 28.50\)
2) \(7x + y = 58.50\)

Subtract equation (1) from equation (2):
\(4x = 30.00\)
\(x = 7.50\)
So, each t-shirt costs £7.50.

Substitute \(x = 7.50\) into equation (1) to find the postage fee:
\(3(7.50) + y = 28.50\)
\(22.50 + y = 28.50\)
\(y = 6.00\)
So, the postage fee is £6.00.

Now, work out the cost of 5 t-shirts including postage:
\(5 \times 7.50 + 6.00 = 37.50 + 6.00 = 43.50\).

M1: for a correct method to find the cost of one t-shirt, e.g. \((58.50 - 28.50) \div 4\) (implied by 7.50).
M1: for a method to find the postage fee, e.g. \(28.50 - 3 \times 7.50\) (implied by 6.00), or a correct expression for 5 t-shirts including postage, e.g. \(5 \times 7.50 + 6.00\).
A1: for 43.50 (accept £43.50).

To calculate the total amount in the account after 3 years, use the compound interest formula:
\(\text{Total Amount} = 4500 \times (1 + 0.024)^3\)
\(\text{Total Amount} = 4500 \times 1.024^3 = 4500 \times 1.073741824 = 4831.838208\)

To find the interest earned, subtract the original investment from the total amount:
\(\text{Interest Earned} = 4831.838208 - 4500 = 331.838208\)

Rounding to the nearest penny gives \(£331.84\).

M1: for a correct method to calculate the total value of the investment after 3 years, e.g. \(4500 \times 1.024^3\) (implied by 4831.84).
M1: for a method to calculate the interest by subtracting 4500 from their total value, e.g. \(4831.838... - 4500\).
A1: for 331.84 (accept £331.84).

First, calculate the total percentage marks of all the girls:
\(18 \times 72 = 1296\)

Next, calculate the total percentage marks of all the boys:
\(12 \times 65 = 780\)

Find the combined total marks of all 30 students:
\(1296 + 780 = 2076\)

Now, divide this total by the number of students (30) to find the mean:
\(\text{Mean} = 2076 \div 30 = 69.2\).

M1: for finding the total marks for the girls, i.e. \(18 \times 72 = 1296\), or the boys, i.e. \(12 \times 65 = 780\).
M1: for a complete method to find the overall mean, i.e. \((1296 + 780) \div 30\).
A1: for 69.2 (accept 69.2%).

The sum of the probabilities for all possible outcomes must equal 1.
Write an equation for the sum:
\(0.15 + 2x + 0.37 + x = 1\)

Simplify the equation:
\(3x + 0.52 = 1\)
\(3x = 1 - 0.52\)
\(3x = 0.48\)
\(x = 0.16\)

Calculate the probability of landing on 2:
\(\text{Probability} = 2x = 2 \times 0.16 = 0.32\)

To find the estimate for the number of times the spinner lands on 2 in 200 spins, multiply the probability by the total number of spins:
\(\text{Estimate} = 200 \times 0.32 = 64\).

M1: for setting up a correct equation for the sum of probabilities, e.g. \(0.15 + 2x + 0.37 + x = 1\), or \(3x = 0.48\).
M1: for finding \(x = 0.16\) and calculating the probability of landing on 2 as \(0.32\), then multiplying by 200.
A1: for 64.

The area of a triangle is given by \(\frac{1}{2} \times \text{base} \times \text{height}\).
Set up the equation using the given values:
\(\frac{1}{2} (2x - 3)(x + 2) = 15\)

Multiply both sides by 2:
\((2x - 3)(x + 2) = 30\)

Expand the brackets:
\(2x^2 + 4x - 3x - 6 = 30\)
\(2x^2 + x - 6 = 30\)

Rearrange into standard quadratic form:
\(2x^2 + x - 36 = 0\)

Factorise the quadratic equation:
\((2x + 9)(x - 4) = 0\)

This gives two solutions for \(x\):
\(x = -4.5\) or \(x = 4\).

Since \(x\) represents lengths, the side length \(2x - 3\) must be positive, which means \(x\) must be positive. Therefore, \(x = 4\).

M1: for setting up the equation for the area, e.g. \(\frac{1}{2}(2x - 3)(x + 2) = 15\) or \(2x^2 + x - 6 = 30\).
M1: for a correct method to solve their quadratic equation \(2x^2 + x - 36 = 0\), e.g. by factorising to \((2x + 9)(x - 4) = 0\) or using the quadratic formula.
A1: for 4 (and rejecting \(-4.5\) as a valid answer).

First, calculate the mass of Liquid A:
\(\text{Mass} = \text{Density} \times \text{Volume}\)
\(\text{Mass of A} = 1.2 \times 250 = 300\text{ g}\)

Next, calculate the mass of Liquid B:
\(\text{Mass of B} = 0.8 \times 150 = 120\text{ g}\)

Calculate the total mass and total volume of Liquid C:
\(\text{Total Mass} = 300 + 120 = 420\text{ g}\)
\(\text{Total Volume} = 250 + 150 = 400\text{ cm}^3\)

Finally, calculate the density of Liquid C:
\(\text{Density of C} = \frac{\text{Total Mass}}{\text{Total Volume}} = \frac{420}{400} = 1.05\text{ g/cm}^3\).

M1: for calculating the mass of Liquid A (\(1.2 \times 250 = 300\text{ g}\)) or the mass of Liquid B (\(0.8 \times 150 = 120\text{ g}\)).
M1: for a complete method to find the density of the mixture by dividing total mass by total volume, e.g. \((300 + 120) \div (250 + 150)\).
A1: for 1.05.

To find the density of the alloy, we need to find the total mass and divide it by the total volume.

First, calculate the mass of Metal A:
\[\text{Mass of A} = \text{Density} \times \text{Volume} = 8.4\text{ g/cm}^3 \times 150\text{ cm}^3 = 1260\text{ g}\]

Next, calculate the mass of Metal B:
\[\text{Mass of B} = \text{Density} \times \text{Volume} = 6.0\text{ g/cm}^3 \times 250\text{ cm}^3 = 1500\text{ g}\]

Now, find the total mass of the alloy:
\[\text{Total mass} = 1260\text{ g} + 1500\text{ g} = 2760\text{ g}\]

Find the total volume of the alloy:
\[\text{Total volume} = 150\text{ cm}^3 + 250\text{ cm}^3 = 400\text{ cm}^3\]

Finally, calculate the density of the alloy:
\[\text{Density} = \frac{\text{Total mass}}{\text{Total volume}} = \frac{2760}{400} = 6.9\text{ g/cm}^3\]

M1: For a method to find the mass of either Metal A or Metal B (e.g. \(8.4 \times 150\) or \(6.0 \times 250\)).
M1: For a complete method to find the density of the alloy, e.g. \(\frac{(8.4 \times 150) + (6.0 \times 250)}{150 + 250}\).
A1: For \(6.9\) (or equivalent).

The perimeter of the garden bed consists of three sides of the rectangle and the curved arc of the semicircle.

First, calculate the length of the curved arc of the semicircle with diameter \(d = 5\text{ m}\):
\[\text{Arc length} = \frac{1}{2} \times \pi \times d = \frac{1}{2} \times \pi \times 5 = 2.5\pi \approx 7.854\text{ m}\]

Next, sum the three straight outer edges of the rectangle (the two lengths of \(12\text{ m}\) and one width of \(5\text{ m}\)):
\[\text{Straight edges} = 12 + 12 + 5 = 29\text{ m}\]

Add these together to get the total perimeter:
\[\text{Total perimeter} = 29 + 7.854 = 36.854\text{ m}\]

Correct to 3 significant figures, this is \(36.9\text{ m}\).

M1: For a correct method to find the arc length of the semicircle, e.g. \(\pi \times 5 \div 2\) or \(\approx 7.85\).
M1: For adding the three sides of the rectangle to their arc length, e.g. \(12 + 12 + 5 + \text{their arc length}\).
A1: For an answer in the range \(36.8\) to \(37.0\) (or \(29 + 2.5\pi\)).

First, form an algebraic expression for the perimeter of the triangle by summing the lengths of its three sides:
\[\text{Perimeter} = (2x + 5) + (2x + 5) + (4x - 3)\]
\[\text{Perimeter} = 8x + 7\]

We are given that the perimeter is \(47\text{ cm}\), so we can set up the equation:
\[8x + 7 = 47\]
\[8x = 40\]
\[x = 5\]

Now, substitute \(x = 5\) back into the side lengths to find their values:
- Side 1: \(2(5) + 5 = 15\text{ cm}\)
- Side 2: \(2(5) + 5 = 15\text{ cm}\)
- Side 3: \(4(5) - 3 = 17\text{ cm}\)

The longest side of the triangle is \(17\text{ cm}\).

M1: For a correct expression for the perimeter, e.g., \((2x + 5) + (2x + 5) + (4x - 3)\) or \(8x + 7\).
M1: (dep) For setting up and solving their equation to find \(x\), e.g., \(8x + 7 = 47\) leading to \(x = 5\).
A1: For \(17\).

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

To find the expected number of times the spinner lands on Yellow, multiply this probability by the total number of spins:
\[\text{Expected frequency} = 0.25 \times 240 = 60\]

M1: For a method to find the probability of landing on Yellow, e.g., \(1 - (0.35 + 0.25 + 0.15)\) or \(0.25\).
M1: For a method to calculate the expected frequency, e.g., \(\text{their } P(\text{Yellow}) \times 240\).
A1: For \(60\).

To find the lower bound of the speed, we use the formula:
\[\text{Speed}_{\text{min}} = \frac{\text{Distance}_{\text{min}}}{\text{Time}_{\text{max}}}\]

First, find the lower bound of the distance:
- Distance is \(400\text{ m}\) to the nearest \(10\text{ m}\), so \(\text{Distance}_{\text{min}} = 395\text{ m}\).

Next, find the upper bound of the time:
- Time is \(52\text{ s}\) to the nearest second, so \(\text{Time}_{\text{max}} = 52.5\text{ s}\).

Now, calculate the lower bound of the average speed:
\[\text{Speed}_{\text{min}} = \frac{395}{52.5} \approx 7.5238... \text{ m/s}\]

Correct to 3 significant figures, this is \(7.52\text{ m/s}\).

M1: For finding either the lower bound of distance (\(395\)) or the upper bound of time (\(52.5\)).
M1: For a division using \(\frac{\text{their } 395}{\text{their } 52.5}\), where \(390 \le \text{their } 395 < 400\) and \(52 < \text{their } 52.5 \le 53\).
A1: For \(7.52\) (or equivalent accurate to 3 sig figs).

To find the value after the first two years at \(3\%\) compound interest:
\[\text{Value after 2 years} = 8000 \times (1.03)^2 = 8000 \times 1.0609 = \pounds 8487.20\]

To find the value after the third year at \(1.5\%\) compound interest:
\[\text{Value after 3 years} = 8487.20 \times 1.015 = \pounds 8614.508\]

Rounding to the nearest penny gives \(\pounds 8614.51\).

M1: For a method to calculate the value of the investment after 2 years, e.g. \(8000 \times 1.03^2\) or \(8487.2(0)\).
M1: For a complete method to find the final value after applying the 3rd year interest, e.g. \(\text{their } 8487.20 \times 1.015\).
A1: For \(8614.51\).

Multiply both sides of the equation by \(5 - p\) to eliminate the fraction:
\[r(5 - p) = 3p + 2\]

Expand the brackets on the left side:
\[5r - rp = 3p + 2\]

Rearrange the equation to collect all terms involving \(p\) on one side and other terms on the opposite side:
\[5r - 2 = 3p + rp\]

Factorise \(p\) out of the right side:
\[5r - 2 = p(3 + r)\]

Divide both sides by \(3 + r\) to isolate \(p\):
\[p = \frac{5r - 2}{3 + r}\]

M1: For a method to clear the fraction, e.g., multiplying both sides by \((5 - p)\) to get \(r(5 - p) = 3p + 2\).
M1: For expanding and gathering terms in \(p\) on one side and constant/non-\(p\) terms on the other, e.g. \(5r - 2 = 3p + rp\).
A1: For \(p = \frac{5r - 2}{3 + r}\) (or equivalent, such as \(p = \frac{2 - 5r}{-3 - r}\)).

1. Find the area of the trapezium cross-section:
\(\text{Area} = \frac{1}{2}(a + b)h = \frac{1}{2}(6 + 10) \times 5 = 8 \times 5 = 40\text{ cm}^2\)

2. Find the volume of one prism:
\(\text{Volume} = \text{Area} \times \text{length} = 40 \times 12 = 480\text{ cm}^3\)

3. Find the total volume of 50 prisms:
\(\text{Total Volume} = 480 \times 50 = 24,000\text{ cm}^3\)

4. Find the total mass of the metal in grams:
\(\text{Mass} = \text{Density} \times \text{Volume} = 7.8 \times 24,000 = 187,200\text{ g}\)

5. Convert the mass to kilograms:
\(\text{Mass in kg} = 187,200 \div 1000 = 187.2\text{ kg}\)

6. Calculate the total cost:
\(\text{Cost} = 187.2 \times \text{£}4.50 = \text{£}842.40\)

M1: for a correct method to find the area of the cross-section, e.g., \(\frac{1}{2}(6 + 10) \times 5\) (\(= 40\))
M1: for finding the volume of 1 prism (\(480\)) or 50 prisms (\(24,000\))
M1: for a correct method to find the total mass using density, e.g., \(24,000 \times 7.8\) (\(= 187,200\))
M1: for converting grams to kilograms (\(\div 1000\)) and multiplying by \(4.50\)
A1: for \(842.40\) (or \(£842.40\), accept \(842.4\) with a loss of the final accuracy mark if not written as currency)

1. Express the areas of the garden and the patio algebraically:
\(\text{Area of garden} = (2x + 5)(x + 3) = 2x^2 + 11x + 15\)
\(\text{Area of patio} = (x - 1)^2 = x^2 - 2x + 1\)

2. Set up the equation for the remaining grass area:
\(\text{Grass Area} = \text{Area of garden} - \text{Area of patio}\)
\(62 = (2x^2 + 11x + 15) - (x^2 - 2x + 1)\)
\(62 = x^2 + 13x + 14\)

3. Rearrange into a standard quadratic equation:
\(x^2 + 13x - 48 = 0\)

4. Factorise the quadratic equation:
\((x + 16)(x - 3) = 0\)
Since dimensions must be positive, \(x\) must be positive, so \(x = 3\) (rejecting \(x = -16\)).

5. Calculate the dimensions of the garden:
\(\text{Length} = 2(3) + 5 = 11\text{ m}\)
\(\text{Width} = 3 + 3 = 6\text{ m}\)

6. Calculate the perimeter:
\(\text{Perimeter} = 2 \times (11 + 6) = 34\text{ m}\)

M1: for expanding \((2x + 5)(x + 3)\) to get \(2x^2 + 11x + 15\) (with at least 3 terms correct) or \((x - 1)^2\) to get \(x^2 - 2x + 1\) (with at least 2 terms correct)
M1: for establishing the equation \((2x^2 + 11x + 15) - (x^2 - 2x + 1) = 62\)
M1: for simplifying the equation to standard quadratic form, e.g., \(x^2 + 13x - 48 = 0\)
M1: for solving the quadratic equation to find \(x = 3\)
A1: for \(34\) (or \(34\text{m}\))

1. Express the probability that both beads are red:
\(\text{P(Red, Red)} = \frac{6}{n} \times \frac{5}{n-1} = \frac{30}{n(n-1)}\)

2. Solve for \(n\):
\(\frac{30}{n(n-1)} = \frac{1}{3} \implies n(n-1) = 90\)
\(n^2 - n - 90 = 0\)
\((n - 10)(n + 9) = 0\)
Since \(n\) must be positive, \(n = 10\).

3. Determine the number of green beads:
There are \(10\) beads in total, so \(10 - 6 = 4\) are green.

4. Calculate the probability of selecting two beads of different colours:
\(\text{P(Red, Green)} = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}\)
\(\text{P(Green, Red)} = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}\)
\(\text{P(Different Colours)} = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}\)

M1: for establishing the algebraic product for \(\text{P(Red, Red)}\), e.g., \(\frac{6}{n} \times \frac{5}{n-1}\)
M1: for creating the equation \(\frac{30}{n(n-1)} = \frac{1}{3}\) or \(n^2 - n - 90 = 0\)
M1: for solving the quadratic equation to find \(n = 10\)
M1: for a complete method to find the probability of different colours, e.g., \(2 \times \left(\frac{6}{10} \times \frac{4}{9}\right)\)
A1: for \(\frac{8}{15}\) (accept equivalent fractions such as \(\frac{48}{90}\) or decimal \(0.533\) or better)

1. Determine the interior angle \(ABC\):
The line from B back to A is at a direction of \(60^\circ + 180^\circ = 240^\circ\).
The bearing of C from B is \(130^\circ\).
Therefore, the interior angle \(ABC = 240^\circ - 130^\circ = 110^\circ\).

2. Use the Cosine Rule to find the distance \(AC\):
\(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)\)
\(AC^2 = 12^2 + 18^2 - 2 \times 12 \times 18 \times \cos(110^\circ)\)
\(AC^2 = 144 + 324 - 432 \times (-0.34202)\)
\(AC^2 = 468 + 147.75 = 615.75\)
\(AC = \sqrt{615.75} \approx 24.814\text{ km}\)

3. Use the Sine Rule to find the angle \(BAC\):
\(\frac{\sin(BAC)}{BC} = \frac{\sin(ABC)}{AC}\)
\(\frac{\sin(BAC)}{18} = \frac{\sin(110^\circ)}{24.814}\)
\(\sin(BAC) = 18 \times \frac{0.93969}{24.814} \approx 0.6816\)
\(\text{Angle } BAC = \sin^{-1}(0.6816) \approx 42.97^\circ\)

4. Calculate the bearing of C from A:
\(\text{Bearing} = 60^\circ + \text{Angle } BAC = 60^\circ + 42.97^\circ = 102.97^\circ\)
To the nearest degree, this is \(103^\circ\).

M1: for calculating the angle \(ABC = 110^\circ\) (may be shown on a sketch)
M1: for applying the Cosine Rule correctly to find \(AC\), e.g., \(12^2 + 18^2 - 2 \times 12 \times 18 \times \cos(110^\circ)\)
A1: for finding \(AC \approx 24.8\text{ km}\) (or value in the range \(24.8\) to \(24.82\))
M1: for applying the Sine Rule (or Cosine Rule) correctly to find the angle \(BAC\), leading to \(BAC \approx 43.0^\circ\)
A1: for \(103\) (accept \(103^\circ\) or answers in the range \(102.9^\circ\) to \(103.1^\circ\))

1. Calculate the compound interest savings account value after 3 years (on 1st January 2023):
\(\text{Savings Value} = 50,000 \times 1.04^3 = 50,000 \times 1.124864 = \text{£}56,243.20\)

2. Determine the machine's value on 1st January 2023:
\(\text{Machine Value} = 56,243.20 - 10,343.20 = \text{£}45,900\)

3. Calculate the machine's value after 2 years of depreciation (end of 2021):
\(\text{Value after Year 1} = 80,000 \times 0.90 = \text{£}72,000\)
\(\text{Value after Year 2} = 72,000 \times 0.85 = \text{£}61,200\)

4. Solve for \(x\) using the third-year depreciation multiplier:
\(61,200 \times \left(1 - \frac{x}{100}\right) = 45,900\)
\(1 - \frac{x}{100} = \frac{45,900}{61,200} = 0.75\)
\(\frac{x}{100} = 0.25 \implies x = 25\)

M1: for a correct method to find the compound interest value, e.g., \(50,000 \times 1.04^3\) (\(= 56,243.20\))
M1: for subtracting \(10,343.20\) from their savings total to find the target machine value (\(45,900\))
M1: for calculating the depreciated value of the machine after 2 years: \(80,000 \times 0.90 \times 0.85\) (\(= 61,200\))
M1: for setting up a correct equation or division for the remaining multiplier: \(\frac{45,900}{61,200}\) (\(= 0.75\))
A1: for \(25\)